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Stoichiometry And Mole Ratio

Mediumchemistry

In the combustion reaction (\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}), how many moles of oxygen are required to completely burn 2.2 g of propane ((\text{C}_3\text{H}_8))? (Molar mass of (\text{C}_3\text{H}_8 = 44) g/mol)

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About This Question

Subject
chemistry
Chapter
some basic concepts in chemistry
Topic
stoichiometry and mole ratio
Difficulty
Medium
Year
2025
Tags
stoichiometrymole ratiocombustionpropanebalanced equation

Solution

Correct Answer:

0.25 mol

Stoichiometry uses the balanced chemical equation to relate moles of reactants and products. The balanced equation shows that 1 mol of propane requires 5 mol of oxygen. First, find the moles of propane: (n_{\text{C}_3\text{H}_8} = 2.2/44 = 0.05) mol. From the mole ratio, moles of (\text{O}_2) required (= 0.05 \times 5 = 0.25) mol. Option 0.05 mol is the number of moles of propane itself, not oxygen — the mole ratio has not been applied. Option 0.50 mol is double the correct value and would correspond to 10 mol of (\text{O}_2) per mol of propane, not 5. Option 1.00 mol is 4 times the correct answer and arises from an incorrect mole ratio of 20:1 instead of 5:1. This is a direct NCERT stoichiometry calculation requiring mole ratio from a balanced equation. Plausibility check: 0.25 mol of (\text{O}_2) has mass (0.25 \times 32 = 8) g, which is reasonable for burning a small amount of propane.

This medium difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of stoichiometry and mole ratio. It appeared in the 2025 exam.

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