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Class 10 Chemistry Chapter 15 Test 6 – Stoichiometry 2027
Grade 10 Chemistry chapter 15 Topic 15.5 to 15.7
Prepare for the Punjab Board Class 10 Chemistry examination with Chapter 15: Stoichiometry Class Test 6, covering Topics 15.5 to 15.7 from the Punjab Board 2027 Model Papers. This Class 10 Chemistry test helps students revise important Stoichiometry concepts, practice relevant questions, and strengthen their understanding of the prescribed topics. Use this Chemistry Chapter 15 Test 6 to improve your exam preparation for the 10th Class annual examination 2027.
Chapter Wise Class Test
T-6
Chapter-15
Stoichiometry
1.
Which of the following is always represented by an empirical formula?
Answer: (c) Ionic compounds
2.
If a compound has 40% carbon, 6.7% hydrogen, and 53.3% oxygen, its empirical formula is:
Answer: (b) CH2O (C : H : O = 3.33 : 6.7 : 3.33 = 1 : 2 : 1)
3.
Which of the following empirical formulae is the simplest form of C6H12O6?
Answer: (b) CH2O (divide all subscripts by 6)
4.
In most experimental chemical reactions, reactants are often used:
Answer: (c) with one or more in excess
5.
The reactant that gets completely consumed first in a reaction is called:
Answer: (c) limiting reactant
6.
What is meant by actual yield?
Answer: (c) Product obtained from experiment
7.
Why is actual yield usually less than theoretical yield?
Answer: (b) Side reactions occur and product is lost
8.
A reaction gives 18 g of product while the theoretical yield is 24 g. What is the efficiency of the reaction?
Answer: (c) 75% (18 ÷ 24 × 100 = 75%)
Chapter Wise Class Test
T-6
Chapter-15
Stoichiometry
2. Write short answers to any six (6) questions.
[6 × 2 = 12]
(i)
Differentiate actual and theoretical yield.
Ans: Theoretical yield is the maximum amount of product calculated from the balanced equation, assuming complete conversion. Actual yield is the amount of product really obtained from the experiment; it is usually less than the theoretical yield.
(ii)
What is relationship between empirical formula and molecular formula.
Ans: Molecular formula = (Empirical formula) n, where n = molecular mass ÷ empirical formula mass. The molecular formula is a whole-number multiple of the empirical formula.
(iii)
Glucose has an empirical formula CH2O and its molecular mass is 180 g mol1−. What will be its molecular formula?
Ans: Empirical formula mass of CH2O = 12 + 2 + 16 = 30
n = 180 ÷ 30 = 6
Molecular formula = (CH2O)6 = C6H12O6
n = 180 ÷ 30 = 6
Molecular formula = (CH2O)6 = C6H12O6
(iv)
Why does the limiting reactant control the amount of product formed?
Ans: The limiting reactant is completely consumed first. Once it is finished the reaction stops, so no more product can form even if other reactants are left over.
(v)
What is percentage purity. How is it calculated?
Ans: Percentage purity tells how much pure substance is present in an impure sample.
% purity = (mass of pure substance ÷ mass of impure sample) × 100
% purity = (mass of pure substance ÷ mass of impure sample) × 100
(vi)
12.0 g sample of a pharmaceutical product was found to contain 11.57 g of the active drug. Find out the percentage purity of the sample.
Ans: % purity = (11.57 ÷ 12.0) × 100 = 96.4 %
(vii)
Why it is necessary for a chemical equation to be balanced before it can be used in calculation?
Ans: A balanced equation obeys the law of conservation of mass and gives the correct mole ratios of reactants and products, which are needed for accurate calculations.
(viii)
Find out the theoretical yield and percent yield of oxygen generated by heating 40 g of KClO3 (M = 122.5). The mass of oxygen gas produced is 14.9g
Ans: 2KClO3 → 2KCl + 3O2
Moles of KClO3 = 40 ÷ 122.5 = 0.3265 mol
Moles of O2 = 0.3265 × 3/2 = 0.4898 mol
Theoretical yield = 0.4898 × 32 = 15.67 g
% yield = (14.9 ÷ 15.67) × 100 = 95.1 %
Moles of KClO3 = 40 ÷ 122.5 = 0.3265 mol
Moles of O2 = 0.3265 × 3/2 = 0.4898 mol
Theoretical yield = 0.4898 × 32 = 15.67 g
% yield = (14.9 ÷ 15.67) × 100 = 95.1 %
(ix)
State why ionic compounds are represented by empirical formulae only.
Ans: Ionic compounds do not exist as discrete molecules. They form a crystal lattice of oppositely charged ions, so only the simplest whole-number ratio of ions can be shown.
Note: Attempt any one question.
[5 × 1 = 5)
3.
Define limiting reactant. How would you identify the limiting reactant in a chemical reaction?
Ans: The reactant that is completely used up first and limits the amount of product is called the limiting reactant.
To identify it:
To identify it:
- Write the balanced chemical equation.
- Convert the given masses of the reactants into moles.
- Divide the moles of each reactant by its coefficient in the equation.
- The reactant with the smallest value is the limiting reactant (or calculate the product from each reactant; the one giving less product is limiting).
4.
Define empirical formula. Write steps to calculate the empirical formula.
Ans: The empirical formula is the simplest whole-number ratio of atoms of the elements present in a compound.
Steps:
Steps:
- Write the percentage (or mass) of each element.
- Divide each percentage by the atomic mass of that element to get moles.
- Divide all the mole values by the smallest value.
- If any ratio is a fraction, multiply all by a suitable number to get whole numbers.
- Write the symbols of the elements with these whole numbers as subscripts.
