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NVAEasyJEE 2023Stoichiometry & Calculations

JEE Chemistry 2023 Question with Solution

Assume carbon burns according to the following equation: 2C(s)+O2(g)→2CO(g)2C(s) + O_2(g) \rightarrow 2CO(g) When 12 g12 \, \text{g} of carbon is burnt in 48 g48 \, \text{g} of oxygen, the volume of carbon monoxide produced is _×10−1\_ \times 10^{-1} L at STP (nearest integer).

Given: Assume CO as ideal gas, Mass of C is 12 g mol−112 \, \text{g mol}^{-1}, Mass of O is 16 g mol−116 \, \text{g mol}^{-1}, and molar volume of an ideal gas at STP is 22.7 L mol−122.7 \, \text{L mol}^{-1}.

Answer

Correct answer:227

Step-by-step solution

Standard Method

Given: Carbon burns according to

2C(s)+O2(g)→2CO(g)2C(s) + O_2(g) \rightarrow 2CO(g)

Mass of carbon = 12 g12 \, \text{g}, mass of oxygen = 48 g48 \, \text{g}, and molar volume at STP = 22.7 L mol−122.7 \, \text{L mol}^{-1}.

Find: The value of the blank in the volume expression of carbon monoxide produced.

First, calculate moles of reactants:

Moles of carbon=12 g12 g mol−1=1\text{Moles of carbon} = \frac{12 \, \text{g}}{12 \, \text{g mol}^{-1}} = 1 Moles of oxygen=48 g32 g mol−1=1.5\text{Moles of oxygen} = \frac{48 \, \text{g}}{32 \, \text{g mol}^{-1}} = 1.5

From the balanced equation,

2C(s)+O2(g)→2CO(g)2C(s) + O_2(g) \rightarrow 2CO(g) 2 mol C:1 mol O2:2 mol CO2 \text{ mol C} : 1 \text{ mol } O_2 : 2 \text{ mol CO}

So, 11 mole carbon requires 0.50.5 mole oxygen and produces 11 mole carbon monoxide.

Since available oxygen is 1.51.5 mol, oxygen is in excess and carbon is the limiting reagent. Therefore, moles of CO produced = 11 mol.

At STP, volume of 11 mol ideal gas is

22.7 L=227×10−1 L22.7 \, \text{L} = 227 \times 10^{-1} \, \text{L}

Hence, the required nearest integer is 227227.

Therefore, the answer is 227227.

Common mistakes

  • Taking oxygen as the limiting reagent is incorrect because 48 g48 \, \text{g} of oxygen corresponds to 1.51.5 mol, while only 0.50.5 mol is needed for 11 mol carbon. Always compare required and available moles before deciding the limiting reagent.

  • Using 16 g mol−116 \, \text{g mol}^{-1} as the molar mass of O2O_2 is wrong. 16 g mol−116 \, \text{g mol}^{-1} is for one oxygen atom, whereas O2O_2 has molar mass 32 g mol−132 \, \text{g mol}^{-1}. Use the molar mass of the actual species present in the equation.

  • Assuming 11 mol carbon gives 22 mol CO is incorrect. From 2C→2CO2C \rightarrow 2CO, the ratio of carbon to carbon monoxide is 1:11:1. Read stoichiometric coefficients proportionally, not in isolation.

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