Vidaara.orgClass 11 · Chemistry
CodeVID-C11-01-T2-01
Stoichiometry & Equations — Practice Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Use atomic masses: H=1, C=12, N=14, O=16, Ca=40.
- Balance every equation before calculating.
- Identify the limiting reagent explicitly where relevant.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The molecular formula is always a whole-number multiple of the:
- A.structural formula
- B.empirical formula
- C.ionic formula
- D.displayed formula
2.
The percentage of carbon in methane ($CH_4$) is:
- A.$25\%$
- B.$50\%$
- C.$75\%$
- D.$80\%$
3.
In $2H_2 + O_2 \rightarrow 2H_2O$, the mole ratio of $H_2$ to $H_2O$ is:
- A.$1:2$
- B.$2:1$
- C.$1:1$
- D.$2:2$
4.
The limiting reagent is the one that:
- A.is in excess
- B.is completely consumed first
- C.is a catalyst
- D.appears as a product
5.
A balanced chemical equation obeys the law of:
- A.definite proportions
- B.conservation of mass
- C.Avogadro
- D.Gay-Lussac
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Balance: $Fe + O_2 \rightarrow Fe_2O_3$.
7.
Calculate the percentage of hydrogen in ammonia, $NH_3$.
8.
Define empirical formula and give the empirical formula of $C_6H_{12}O_6$.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
An organic compound has 92.3% carbon and 7.7% hydrogen and a molar mass of 78 g/mol. Find its molecular formula.
10.
How many grams of $CO_2$ are produced by burning $16\,g$ of methane, $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$?
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
$28\,g$ of $N_2$ is mixed with $6\,g$ of $H_2$ for $N_2 + 3H_2 \rightarrow 2NH_3$. Find the limiting reagent, mass of $NH_3$ formed and mass of excess reagent left.
Answer Key
Section A — Multiple Choice Questions
- (B) empirical formula
- (C) $75\%$
- (C) $1:1$
- (B) is completely consumed first
- (B) conservation of mass
Section B — Short Answer (2 marks)
- 4Fe + 3O2 -> 2Fe2O3.
- M = 17; H mass = 3; % H = (3/17) x 100 = 17.6%.
- Empirical formula is the simplest whole-number ratio of atoms. For C6H12O6 it is CH2O.
Section C — Short Answer (3 marks)
- C = 92.3/12 = 7.69; H = 7.7/1 = 7.7; ratio 1:1 so empirical formula CH (mass 13). n = 78/13 = 6, molecular formula C6H6.
- n(CH4) = 16/16 = 1 mol; gives 1 mol CO2 = 44 g.
Section D — Long Answer (5 marks)
- n(N2) = 28/28 = 1 mol; n(H2) = 6/2 = 3 mol. Per coefficient: N2 = 1/1 = 1, H2 = 3/3 = 1, both equal, so reactants are in exact ratio with no excess. NH3 formed = 2 mol = 2 x 17 = 34 g; no reactant left over.
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