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10th Class - Physical Science-TS

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10th Class - Physical Science-TS
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I. Reflections on concepts

Question 1.
What information do you get from a balanced chemical equation?

Answer:

  1. A chemical equation gives information about the reactants and products by means of their symbols and formulae.
  2. It gives the ratio of molecules of reactants and products.
  3. It gives the relative masses of reactants and products.
  4. If the masses are expressed in grams, then the equation also gives the molar ratios of reactants and products.
  5. We can calculate the volumes of gases liberated at given condition of temperature and pressure using molar mass and molar volume relationship.
  6. Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances.
Question 2.
Why should we balance a chemical equation?

Answer:
Chemical reactions obeys law of conservation of mass. So, the total number of atoms of each element in the reactants must be equal to the total number of atoms of each element in the products. So we should have to balance chemical equation.

Question 3.
Balance the following chemical equations.

(a) NaOH + H2SO4→ Na2SO4+ H2O

Answer:
Step 1: Write unbalanced equation
NaOH + H2SO4→ Na2SO4+ H2O

Step 2: Compare number of atoms of each element on both sides. Add the suitable coefficients to balance equation.
2NaOH + H2SO4→ Na2SO4+ 2H2O

Step 3 : Make sure the coefficients are reduced to their smallest whole number values.
2NaOH + H2SO4→ Na2SO4+ 2H2O

Step 4: Check the answer

(b) KClO3→ KCl + O2
Answer:
Step 1: Write unbalanced equation.
KClO3→ KCl +O2

Step 2: Add suitable coefficients to balance equation on both sides.
2KClO3→ 2KCl +3O2

Step 3: Make sure the coefficients are reduced to their smallest whole number values.
2KClO3→ 2KCl +3O2

Step 4: Check the answer.

(c) Hg (NO3)2+ Kl → Hg I2+ KNO3
Answer:
Step 1: Write unbalanced equation
Hg(NO3)2+KI → HgI2+KNO3

Step 2: Add suitable coefficients to balance equation on both sides
Hg(NO3)2+2KI → HgI2+2KNO3

Step 3: Make sure the coefficients are reduced to their smallest whoíe number values.
Hg(NO3)2+2 Kl → Hg I2+ 2KNO3

Step 4: Check the answer

Question 4.
Mention the physical states of the reactants and products of the following chemical reactions and balance the equations.

(a) C6H12O6→ C2H5OH + CO2

Answer:
C6H12O6(aq) → 2C2H5OH(aq)+2CO2(g)

(b) NH3+ Cl2→ N2+ NH4Cl(s)
Answer:
8NH3(g)+3Cl2(g)→ N2(g)+6NH4Cl

(c) Na+H2O → NaOH+H2
Answer:
2Na(s)+2H2O (l) → 2NaOH (aq)+H2(g)

II. Application Of Concepts

Question 1.
Balance the following chemical equations after writing the symbolic representation.

(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)
(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)

Answer:
(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)

Answer:
Ca(OH)2(sol)+ 2HNO3(so1)→ 2H2O(l)+ Ca(NO3)2(aq)Double displacement reaction

(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)

Answer:
Mg(s)+ I2(s)→ MgI2(s)
Chemical combination reaction.

Question 2.
Write the following chemical reactions including the physical states of the substances and balance chemical equations

(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.

Answer:
(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.

Answer:
NaOH (aq) + HCl (aq) ) → NaCl (aq) +H2O (l)

(b) Barium chloride reacts with liquid sodium sulphate to leave Barium sulphate as a precipitate and also form liquid sodium chloride.

Answer:
BaCl2(aq)+ Na2SO4(aq)→ BaSO4↓ +2 NaCl(aq )

Question 3.
Potassium nitrate and Sodium nitrate reacts separately with copper sulphate solution. Write balanced chemical equations for the above reactions.

Answer:

  1. Potassium nitrate reacts with copper sulphate to form potassium sulphate and copper nitrate.
    2KNO3+ CuSO4→ K2SO4+ Cu(NO3)2
    2. Sodium nitrate reacts with copper sulphate to form sodium sulphate and copper nitrate.
    2NaNO3+ CuSO4→ Na2SO4+ Cu(NO3)2

Higher Order Thinking Questions

Question 1.
2 moles of zinc reacts with a cupric chloride solution containing 6.023 × 1022formula units of CuC12. Calculate the moles of copper obtained.
Zn(s) + CuCl2(aq)→ ZnCl2(aq)+ Cu (s)

Answer:
Given equation i.s
Zn(s) + CuCl2(aq)→ ZnCl2(aq)+ Cu(s)
1 mol + 1 mole → 1 mole + mole
From the above equation it is clear that 1 mole of zinc react with 1 mole of CuCl2solution to give 1 mole of copper.
2 moIes of Zn required 2 moles (12.046 × 1022formula units) of CuCl2, But only 1 mole (6.023 × 1022formula units) of CuCl2is available.
So, the No. of moles of copper obtained depends on amount of CuCl2present.
∴6.023 × 1022formula units (1 mole) of CuCl2products 1 mole of copper.

Question 2.
1 mole of propane (C3H8) on combustion at STP gives A’ kilo Joules of heat energy. Calculate the heat liberated when 2.4 ltrs of propane on combustion at STP.

Answer:
The chemical equation for combination of propane is
C3H8+ 5O2→ 3CO2+ 4H2O + A (heat energy)
1 mole of propane gives ‘A kilojoules of heat energy.
(1 mole of any gas occupies 22.4 litres at SW)
i.e., 22.4 ltrs. of propane gives ‘A’ kilojoules of heat energy
⇒ 2.25 ltrs. of propane gives 2.24/22.4 x A = 1/10 A (0.1 A) heat energy.

Question 3.
Calculate the mass and volume of Oxygen required at STP to convert 2.4 kg of graphite Into carbon dioxide.

Answer:
The chemical equation is
C + O2→ CO2(∴ graphite is also carbon)
12 gm + 32 gm → 44 gm
1 mole + 1 mole → 1 mole
From the above equation
12 gm of Graphite requires 32 gm of oxygen

1 mole of oxygen occupies 22.4 litres
200 moles Is oxygen occupIes 22.4 x 200 4480 litres.

Page 20

Question 1.
How do we know a chemical reaction has taken place?

Answer:
a. The original substances lose their characteristic properties. Hence these may be products with different physical states and colours permanently.

b. Chemical changes may be exothermic or endothermic i.e. they may involve liberation of heat energy or absorption of heat energy.

c. They may form an insoluble substance known as precipitate.

d. There may be gas liberation in a chemical change.

Page 21

Activity 1.
Take about 1 g of quick lime (calcium oxide) in a beaker. Add 10 ml of water to this.
Touch the beaker with your finger.

Question 2.
What do you notice?

Answer:
We notice that the beaker is hot when we touch it. The reason is that the calcium oxide (quick lime) reacts with water and in the process heat energy Is released. Calcium oxide dissolves in water producing colourless solution of Ca(OH)2.

Question 3.
What is the nature of the solution?

Answer:
This solution Is a basic solution because a red litmus paper turns blue when dipped in the above solution

Activity 2
Take about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na2SO4).
Take about 100ml of water In another beaker and dissolve a small quantity of barium chloride (BaCl2) observe the colours of the solutions obtained.

Question 4.
What are the colours of the above solutions?

Answer:
Colourless solutions

Question 5.
Can you name the solutions obtained?

Answer:
Sodium sulphate solution and barium chloride solution

Question 6.
Add Na2SO4solution to BaCl2solution and observe. Do you observe any change on mixing these solutions?

Answer:
A white precipitate of barium sulphate is formed

Activity 3
Take a few zinc granules in a conical flask. Add about 5 ml of dilute hydrochloric acid to the conical flask.

Question 7.
What changes do you notice?

Answer:
Effervescence Is observed

Question 8.
Keep a burning match stick near the mouth of the conical flask. with happens to burning match stick?

Answer:
It is put out. A pop sound is also heard.

Question 9.
Touch the bottom of the conical flask with your fingers. What do you notice?

Answer:
It Is hot

Question 10.
Is there any change in temperature?

Answer:
Temperature increases.

Page 22

Question 11.
Can you write a chemical reaction in any other shorter way other than the way we disused above?

Answer:
The reaction of calcium oxide with water can be written using symbols of elements.
CaO + H2O → Ca(OH)2

Page 23

Question 12.
Is the number of atoms of each element are equal on both sides?

Answer:
The number of atoms of each element on both sides is equal.

Question 13.
Sodium sulphate reacts with barium chloride to give white precipitate, barium sulphate.
Na2SO4+ BaCl2→ BaSO4↓ + NaCl

Question 14.
Do the atoms of each element on left side equal to the atoms of the elements on the right side of the equation?

Answer:
No, the sodium and chlorine atoms are not balanced.

Page 25

Question 15.
Is it a balanced equation as per rules?

Answer:
Yes. But the coefficients are not the smallest numbers.

Question 16.
How do you say?

Answer:
Though the equation is balanced, the coefficients are not the smallest whole numbers. It would be necessary to divide all coefficients of equation by 2 to reach the final equation.
C3H8+ 5O2→ 3CO2+ 4H2O

Think And Discuss
You have brushed the wall with an aqueous suspension of Ca(OH)2. After two days the wall turned to white colour.

Question 1.
What are the steps involved in whitewashing of walls? (P.No. 23)

Answer:
A solution of slaked lime [Ca(OH)2] is prepared by adding water to quick lime [CaoO]. When Ca(OH)2is applied to the wall ¡t reacts with carbon dioxide in air to form a thin layer of calcium carbonate giving a shiny finish to the walls.

Question 2.
Write the balanced chemical reactions using the appropriate symbols. (P.No. 23)

Answer:
CaO(s)+ H2O(l)→Ca(OH)2(Aq)+ Q(heat energy)

Activity 1

Question 1.
How can you demonstrate action of quick lime with water? What is the nature of the product? (2 Marks)

Answer:

  1. Take about 1 g of quick lime (Calcium oxide) in a beaker. Add 10 ml of water to this. Touch the beaker with your fingers
  2. We will notice that the beaker is hot when we touch it. Hence that’s an exothermic reaction.
  3. The reason is that the Calcium oxide reacts with water and in that process heat energy is released.
  4. Calcium oxide dissolves In water producing colourless solution, This solution turns red litmus to blue. Hence the product is a base.
    CaO + H2O → Ca(OH)2+ Q

Activity 2

Question 2.
Explain the reaction between Sodium sulphate and Barium chloride.

Answer:

  1. luke about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na2SO4)
  2. Take about 100 ml of water in another beaker and dissolve a small quantity of Barium chloride (BaCl2).
  3. These two solutions are colourless.
  4. Add Na2SO4solution to BaCl2solution and observe.
  5. Sodium sulphate solution on mixing with Barium chloride solution alarms a precipitate of Barium sulphate and also soluble Sodium chloride.
    Na2SO4(aq)+ BaCl(aq)→ BaSO4(s)↓ + 2NaCl(aq)
  6. This is a double displacement reaction.
Question 3.
Explain the reaction of Zinc with HCl and write a balanced equation.
(OR)
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’pieces and liberates H2”.

Answer:
Material required :
(i) Conical flask
(ii) Zinc granules
(iii) HCl
(iv) Matchbox.

  1. Take a few zinc granules in a conical flask.
  2. Add about 5 ml of dilute Hydrochloric acid to the conical flask.
  3. We observe a gas evolving out.
  4. Now keep a burning match stick near the mouth of the conical flask.
  5. The match stick is put out with a ‘pop’ sound.
  6. I Touch the bottom of the conical flask. We feel hot. Hence it is an exothermic reaction.

Zn(s) + 2HCl(l) → ZnCl2(s) + H2(g) ↑

Important Question

TS 10th Class Physical Science Important Questions Chapter 2 Chemical Equations

1 Mark Questions
Question 1.
What are the reactants and products in a chemical reaction?

Answer: The substances which undergo chemical changes In a reaction are called reactants and the new substances formed are called products

Question 2.
What Is meant by balanced equation?

Answer: A chemical equation Is said to be balanced when the number of atoms of each element is same on both reactant side and product side

Question 3.
If you keep an iron piece in solid-state CuSO4 crystals, does it get any reaction? Guess the reason?

Answer: Reaction will not takes place if an ¡ron piece is placed in solid state CuSO4 crystals because there will not exist separate Cu2+, SO4-2 ions in CuSO4 crystals. In aqueous solution, they exists. So iron cannot displace copper

Question 4.
Baiancethefollowlng equations?
  1. Na + O2 ? Na2O
  2. H2O2 ? H2O + O2
  3. Mg(OH)2 + HCl ? MgCl2 + H2O
  4. Fe + O2 ? Fe2O3

Answer:

  1. 4Na + O2 ? 2Na2O
  2. 2H2O2 ? 2H2O + O2
  3. Mg(OH)2+ 2HCl ? MgCl2 + 2H2O
  4. 4Fe + 3O2 ? 2Fe2O3
Question 5.
Balance the following equations. (AS1)?
  1. Al(OH)3 ? Al2O3 + H2O
  2. NH3 + CuO ? Cu + N2 + H2O
  3. Al2(SO4)3 + NaOH ? Al(OH)3 + Na2SO4
  4. HNO3 + Ca(OH)2 ? Ca(NO3)2 + H2O
  5. NaOH + H2SO4 ? Na2SO4 + H2O
  6. BaCl2 + H2SO4 ? BaSO4 + HCl

Answer:

  1. 2Al(OH)3 ? Al2O3 + 3H2O
  2. 2NH3 + 3CuO ? 3Cu + N2 + 3H2O
  3. Al2(SO4)3+ 6NaOH ? 2Al(OH)3 + 3Na2SO4
  4. 2HNO3 + Ca(OH)2 ? Ca(NO3)2 + 2H2O
  5. 2NaOH + H2SO4 ? Na2SO4 + 2H2O
  6. BaCl2 + H2SO4 ? BaSO4 + 2HCl
Question 6.
What is a chemical equation?

Answer: Describing a chemical reaction using least possible words or symbols is called a chemical equation

Ex : CaO + H2O ? Ca(OH)2
Question 7.
MnO2 + 4HCl ? MnCl2+ 2H2O + Cl2. In the given equation, name the compound which is oxidized and which is reduced?

Answer: In the above equation Ha compound is oxidized and MnO2 is reduced

Question 8.
What happens If the copper sulphate crystals taken into dry test tube are heated?

Answer:

  1. When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white
  2. Evaporated water appears as droplets on the walls of the test tube
  3. Blue-coloured copper sulphate (CuSO45H2O) is turned into white colour beacuse 5H2O molecules are evaporated from crystals
Question 9.
Write the equation for the chemical decomposition reaction of silver chloride In the presence of sunlight?

Answer:

2 Marks Questions
Question 1.
What do you observe during chemical change?

Answer: During chemical change, we observe that

  1. The original substances have their characteristic properties. Hence these may be products with different physical states colours and different compositions.
  2. Chemical changes may be exothermic or endothermic
  3. They may form an Insoluble substance known as precipitate
  4. There may be gas liberation in a chemical danger
Question 2.
How do you Write a chemical equation?

Answer:

  1. A chemical reaction, written tri the form of formula equation, shows the change of reactants to products by an arrow placed between them
  2. The reactants are written on the left side of arrow and the final substances or products are written on the right side of the arrow
  3. The arrowhead points towards the products and shows the direction of reaction
  4. If there are more than one reactant or product involved in the reaction, they are separated with a plus (+) sign between them
Question 3.
What are the effects of oxidation on everyday life?

Answer:

  1. Combustion is the most common example for oxidation reactions
  2. Rising of dough with yeast depends on oxidation of sugars to carbon dioxide and water
  3. Metals are corroded due to oxidation
  4. Bleaching of coloured objects using moist chlorine. Cl + H2O ? HOCl + HCl HOCl ? HCl + (O) Coloured object + (0) ? Colourless object (Here O is known as Nascent oxygen)
  5. Sometimes during rainy season the power supply to our homes from the electric pole will be interrupted due to formation of metal oxide layer on the electric wire. This metal oxide is an electrical insulator. On removing the metal oxide layer formed on the wire with sandpaper, supply of electricity can be restored
Question 4.
What are the important characteristics of chemical reactions?

Answer: The important characteristics of chemical reactions are

  1. Evolution of a gas
  2. Formation of a precipitate
  3. Formation of new substances
  4. Change in colour
  5. Change in temperature
  6. Change in state
Question 5.
Write some chemical reactions occurring in our daily life?

Answer:

  1. Souring of milk
  2. Formation of curd from milk
  3. Cooking of food
  4. Digestion of food in our body
  5. Fermentation of grapes
  6. Rusting of iron
  7. Burning of fuels
  8. Burning of cooking gas
  9. Ripening of fruits
Question 6.
What symbols do we use to indicate the physical state of reactants and products In an equation?

Answer:

  1. Solid state is indicated by the symbol (s)
  2. Liquid state is indicated by the symbol ( l )
  3. Gaseous state is indicated by the symbol (g)
  4. Aqueous solution is indicated by the symbol (aq)
  5. Precipitate Is Indicated by an arrow with head downwards (?)
Question 7.
Comment on "C(s) + O2(g) ? CO2(g) + Heat equation?

Answer:

  1. The burning of carbon In oxygen Is an exothermic reaction because heat is evolved in this reaction
  2. An exothermic reaction is indicated by writing + Heat or + Heat energy or just+ Energy on product side of an equation
Question 8.
Comment on N2(g) + O2(g) + Heat ? 2NO(g)" equation?

Answer:

  1. The reaction between nitrogen and oxygen to form Nitrogen Monoxide is endothermic reaction because heat is absorbed n this reaction
  2. An endothermic reaction is usually indicated by writing + Heat or Just + Energy on the reactants side of an equation
Question 9.
2Cu + O2 ? 2CuO What information do you get from above equation?

Answer: The above equation tells us that

  1. Copper reacts with oxygen to form copper oxide
  2. The formula of copper oxide.is, CuO and that of oxygen is
  3. 2 moles of copper atoms react with 1 mole of oxygen molecules (O2) to produce 2 moles of copper oxide (CuO)
Question 10.
Balance the following chemical equation and follow the steps involved in balancing a chemical equation?
Cu2S + O2 ? Cu2O + SO2

Answer: Step 1: Write the unbalanced equation using correct chemical formula for all substances Cu2S + O2 ? Cu2O + SO2 Step 2: Compare no. of atoms of each element on both sides

Atom No. of atoms in L.H.S No. of atoms in R.H.S.
Cu 2 2
S 1 1
0 2 3

Balancing Cu, S, O atoms both sides 2Cu2S + 3O2 ? 2Cu2O + 2SO2 The equation is balanced. Step 3: Write the co-efficient smallest ratio. 2Cu2S +3O2 ? 2Cu2O + 2SO2 Step 4: Verify above equation for balancing of atom of each element on both sides. 2Cu2S + 3O2 ? 2Cu2O + 2SO2

Question 11.
Observe the following balanced chemical equation and answer the given
Questions. C3H8(8) + 5O2(g) ? 3CO2(g) + 4H2O(g)?
  1. How many molecules of Oxygen are involved In this chemical reaction
  2. How many moles of Propane are required to get 20 moles of Water

Answer:

  1. In this chemical reactIon five molecules of oxygen are involved
  2. Five moles of propane are required to get 20 moles of water
Question 12.
Write the products of given reactions, if any. Give reason?
FeCl2 + Zn ? ZnCl2 + Fe ?

Answer: FeCl2 + Zn ? ZnCl2+Fe (Displacement reaction) ZnCl2 + Fe ? No reaction. (Low reactive metals cannot displace high reactive metals)

Question 13.
Balance the following chemical equations?
  1. Na+ H2O ? NaOH+H2
  2. K2CO3+HCl ?KCl+H2O+CO2

Answer:

  1. 2Na + 2H2O ? 2NaOH + H2
  2. K2CO3 +2HCl ? 2KCl + H2O+CO2
4 Marks Questions
Question 1.
What are the materials required for the experiment to show the chemical decomposition of water? Write the procedure of the experiment. Name the products which we get in this reaction?

Answer: Material required for chemical decomposition of water: A plastic mug, two carbon rods, two corks, two test tubes, connecting wires, 9V battery, water, and some drops of acid

Procedure:
  1. Take a plastic mug, drill holes at the base
  2. Fit two one-holed rubber stoppers in these holes
  3. Insert two carbon electrodes in these rubber stoppers
  4. Connect the electrodes to 9V battery
  5. Fill the mug with water, so that the electrodes are immersed.
  6. Add few mops of any acid
  7. Take two test tubes tilled with water and insert them over the two carbon electrodes
  8. Switch or the circuit and leave the apparatus undisturbed for some time
  9. We notice the liberation of bubbles at both the electrodes. These bubbles displace the water in the test tubes
  10. After the test tubes are filled with gas, take them out and test the gases with the burning match Stick

The products which we get in this reaction: The gases are condensed oxygen and hydrogen

Question 2.
Explain the steps involved in balancing a chemical equation with an example?
(or) Why should we balance a chemical equation? Take any one chemical equation and explain the procedure of balancing it.

Answer: A chemical equation in which the number of atoms of different elements on the reactants side are same as those on product side is called a balanced equation. Steps involved In balancing a chemical reaction: Let us consider the combustion reaction of Propane. Step 1: Write the unbalanced equation using correct chemical formulae for all substances. C3H8 + O2 ? CO2 + H2O (Skeleton equation) Step 2: Compare number of atoms of each element on both sides

Find the coefficients to balance the equation. In this case, there are 3 carbon atoms on the left side of the equation but only one on the right side. If we add a coefficient of 3 to CO2 on the right side the carbon atoms balance. C3H8 + O2 ? 3CO2 + H2O Now, look at the number of hydrogen atoms. There are 8 hydrogen atoms on the left but only 2 on the right side. By adding a coefficient of 4 to the H2O on the right side, the hydrogen atoms get balanced. C3H8+ O2 ? 3CO2 + 4H2O Finally, look at the number of oxygen atoms. There are 2 on the left side but 10 on the right side. By adding a coefficient of 5 to the 02 on the left side, the oxygen atoms get balanced. C3H8 + 5O2 ? 3CO2 + 4H2O

Step 3: Make sure the coefficients are reduced to their smallest whole number values. The above equation is already with the coefficients in smallest whole numbers. There is no need to reduce its coefficients. Hence the final equation is C3H8 + 5O2 ? 3CO2 + 4H2O

Step 4: Check the answer. Count the numbers and kinds of atoms on both sides of the equation to make sure they are the same

Question 3.
How to make a chemical equation more informative?

Answer: Chemical equations can be made more informative by expressing the following characteristics of the reactants and products

  1. Physical state
  2. Heat exchange (exothermic or endothermic change)
  3. Gas evolved (if any)
  4. Precipitate formed (If any)
  5. Expressing the physical state: The physical state of the substances maybe mentioned along with their chemical formulae. The different states i e., gaseous, liquid and solid states are represented by the notations(g), (I) and (s) respectively. If the substance is present as a solution in water, it Is represented as (aq). Eg : Fe2O3(s) + 2Al(s) ? 2Fe(s) +Al2O< sub>3(s)
  6. Heat exchange: Heat is liberated in exothermic reactions and heat is absorbed in endothermic reactions. Eg: 1. C(s) + O2(g) ? CO2(g) + Q (exothermic reaction) 2. N2(g)+ O2(g) ? 2NO(g) - Q (endothermic reaction) (or) 
  7. Gas evolved: If a gas is evolved in a reaction, it is denoted by an upward arrow ?or (g) Eg: Zn(s) + H2SO4(aq) ? ZnSO4(aq) + H2(g) ?
  8. Precipitate formed: If a precipitate is formed in the reaction, it Is denoted by a downward arrow. Eg : AgNO3(aq) + NaCl(aq) ? AgCl(s) ? + NaNO3(aq)
Question 4.
Balance the following chemical equations?
  1. Na2SO4 + BaCl2 ? BaSO4 + NaCl
  2. Al4C3 + H2O ? CH4 + Al(OH)3
  3. Pb(NO3)2 ? PbO + NO2 + O2
  4. Fe2O3 + Al ? Al2O3 + Fe

Answer:

  1. Na2SO4 + BaCl2 ? BaSO4 + 2NaCl
  2. Al4C3 + 12H2O ? CH4 + 4Al(OH)3
  3. 2Pb(NO3)2 ? 2PbO + 4NO2 + O2
  4. Fe2O3 + 2Al ? Al2O3 + 2Fe
Solved Problems
Question 1.
Al(s) + Fe2O3(s) ? Al2O3(s) + Fe(s) (atomic masses of Al = 27U, Fe = 56U, and O = 16U) Suppose that you are asked to calculate the amount of aluminum, required to get 1120 kg of iron by the above reaction?

Answer: As per the balanced equation Aluminium ? Iron 54g ? 112g X? ? (1120 x 1000)g ? xg = (1120×1000)g×54112 g = 10000 x 54g = 540000 g (or) 540 kg. To get 1120 kg of iron we have to use 540 kg of aluminum

Question 2.
Calculate the volume, mass, and number of molecules of hydrogen liberated when 230 g of sodium reacts with excess of water at STP. (atomic masses of Na = 23U), O=16U, and H=1U) The balanced equation for the above reaction is?

Answer: As per the balanced equation 46g of Na gives 2g of hydro9en. 230g of Na gives .......? g of hydrogen. 230 g×2 g46 g = 10g of hydrogen. I gram molar mass of any gas at STP ie, standard temperature 273K and standard pressure 1 bar occupies 22.4 liters known as gram molar volume. 2.0 g of hydrogen occupies 22.4 liters at STP. 10.0 g of hydrogen occupies.. ? litres at STP. 10.0 g×22.4 litres 2.0 g = 112 litres 2 g of hydrogen i.e, 1 mole of H2 contains 6.02 x 1023 (H2) molecules 10 g of hydrogen contain ......... ? 10.0 g×6.02×1023 molecules 2.0 g = 30.10 x 1023 molecules = 3.01 × 1024 molecules

Question 3.
Calculate the volume and No. of molecules of CO2 liberated at STP if 50g of CaCO3 is treated with dilute hydrochloric acid which contains 7.3 g of dissolved HCI gas?

Answer: The Chemical equation for the above the relation is CaCO3(s) + 2HCl (aq) ? CaCl2(aq) + H2O(l) + CO2(g) As per the stoichiometric equation, 100 g of CaCO3 reacts with 73 kg of HCl to liberate 44 g of CO2. In the above problem the amount of CaCO3, taken is 50 gm and HCl available is 7.3 g 100g of CaCo3, require 73g of HCl and 50 g of CaCo3 required 36.6 g of HCl but, only 7.3 g of HCI is available. Hence the product CO2 formed depends only on the amount of HCl which Is in the the least amount but not on the amount of CaCO3 which is an excess. The reactant available in less amount Is called limiting reagent as it limits the amount of prouct formed. Therefore, we can write 73 g of HCl ? 44 g CO2 7.3g of HCl- ? 7.3g×44g73g = 4.4 g 44 g of CO2 occupies 22.4 L volume at STP 4.4 g of CO2 occupies - ? 4.4g×22.4L44g = 2,24L 44 g of CO2 contain 6.023 × 1023 mol of CO2 4.4 g Contains - ? 4.4g×6.023×102344g = 6.023 × 1022 mol

Question 4.
Write the equation for the reaction of Zinc with hydrochloric acid and balance the equation. Find out the number of molecules of hydrogen gas produced In this reaction, when 1 mole of HCl completely reacts at S.T.P. [Gram molar volume is 22.4 liters at S.T.P., Avogadros number is 6.023 x1023]?

Answer: Zn + HCl ? ZnCl2+H2 Zn(s) + 2HCl(l) ? ZnCl2(s)+H2(g) From the above equation 2 moles of HCl produces 1 moIe of H2 gas 1 moIe of HCl produces 12 mole of H2 gas 1 mole of H2 gas at S.T.P Contains 6.023 x 1023 molecules 12 mole of H2 gas contains = 6.023×10232 = 3.0115 x 1023 molecules

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