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Class 10 Chemistry Chapter 15 Stoichiometry Test 4 – 2027 | Solved Paper

Grade 10 Chemistry chapter 15 Test 4 with answers

Prepare for the Punjab Board Class 10 Chemistry examination with the Chapter 15: Stoichiometry Class Test 4 from the Punjab Board 2027 Model Papers. This test is designed to help students revise important Stoichiometry concepts, chemical equations, numerical problems, and related Chemistry questions. Practice this Class 10 Chemistry Chapter 15 Test 4 to strengthen your conceptual understanding and improve your preparation for the Punjab Board annual examination 2027.

Chapter: 15
Total Time: 35 min

CHAPTER WISE CLASS TEST

(Test # 4)

Chemistry
Total Marks: 25
ABCD
1ABCD
2ABCD
3ABCD
ABCD
4ABCD
5ABCD
6ABCD
ABCD
7ABCD
8ABCD
NOTE: Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
TIME: 10 Min
OBJECTIVE
MARKS: 8
Q.1:(08)
1.

Find out the concentration of H2SO4 in mol/dm3 if its 10 cm3 react with 20 cm3 of NaOH solution having concentration 0.1 mol/dm3.

(A) 0.01 mol/dm3 (B) 0.001 mol/dm3 (C) 0.1 mol/dm3 (D) 0.15 mol/dm3
H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles NaOH = 0.1 × 0.020 = 0.002 mol; moles H2SO4 = 0.001 mol in 0.010 dm3 ⇒ 0.1 mol/dm3.
2.

The empirical formula of magnesium chloride is:

(A) MgCl (B) MgCl2 (C) Mg2Cl (D) Mg2Cl2
3.

If the molar mass of a compound is 178 g/mol and its empirical formula is C3H5O3. What is its molecular formula?

(A) C3H5O3 (B) C6H10O6 (C) C9H15O9 (D) C8H10O8
Empirical formula mass = 89; n = 178 ÷ 89 = 2 ⇒ C6H10O6.
4.

The reactant which is consumed completely in a chemical reaction is called:

(A) Excess reactant (B) Limiting reactant (C) Product (D) Catalyst
5.

When natural gas burns in air, which component is the limiting reactant?

(A) Both components (B) Neither component is a limiting reactant (C) Air (D) Natural gas
6.

The theoretical yield of Zn(NO3)2 from 0.76 mol of Zn is:

(A) 100.00 g (B) 120.50 g (C) 143.92 g (D) 160.00 g
Zn → Zn(NO3)2 (1:1). Mass = 0.76 mol × 189.4 g/mol ≈ 143.92 g.
7.

If the actual yield of a compound is 0.198 g while the theoretical yield is 0.217 g. What will be the percentage yield?

(A) 50% (B) 91.2% (C) 80% (D) 90
(0.198 ÷ 0.217) × 100 = 91.2%.
8.

The formula to calculate percentage yield is:

(A) Theoretical yield ÷ Actual yield × 100 (B) Actual yield ÷ Theoretical yield × 100 (C) Actual yield × Theoretical yield × 100 (D) Actual yield + Theoretical yield × 100
TIME: 25 Min
SUBJECTIVE
MARKS: 17

(PART – I)

Q.2Write short answers to following questions.(6×2=12)
  1. Why is percentage yield important?
    It shows the efficiency of a reaction by comparing actual yield with theoretical yield, helping chemists and industries judge cost, waste and the best conditions for a process.
  2. How can molecular formula of a compound be calculated?
    Find the empirical formula mass, divide the given molar mass by it to get n, then multiply every subscript of the empirical formula by n: Molecular formula = (Empirical formula)n.
  3. Will there be a limiting reactant in a reversible reaction?
    No. In a reversible reaction the reaction does not go to completion, so no reactant is consumed completely and a true limiting reactant does not arise.
  4. How does the limiting reactant control the amount of product formed?
    The reaction stops when the limiting reactant is used up, so the amount of product is fixed by the moles of the limiting reactant according to the balanced equation.
  5. How do you know if a formula is empirical or not?
    If the subscripts of all atoms have no common factor (simplest whole-number ratio), it is empirical. If they can all be divided by a common number, it is not (e.g. C6H12O6 → CH2O).
  6. What is percentage yield and write its formula?
    Percentage yield is the ratio of actual yield to theoretical yield expressed as a percentage.
    % Yield = (Actual yield ÷ Theoretical yield) × 100

(PART – II)

Note: Attempt the following question.
Q.3: How is chemical equation helpful in calculations?(5)
A balanced chemical equation gives:
  • the mole ratio of reactants and products (e.g. 2H2 + O2 → 2H2O);
  • the masses of reactants needed or products formed, using molar masses;
  • volumes of gases at STP (1 mol = 22.4 dm3);
  • the limiting reactant, and hence theoretical and percentage yield;
  • the number of particles involved (1 mol = 6.02 × 1023).

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