Question

Assume carbon burns according to following equation : 
$2 C _{( s )}+ O _{2( g )} \rightarrow 2 CO ( g )$ 
When $12 g$ carbon is burnt in $48 g$ of oxygen, the volume of carbon monoxide produced is___ $\times 10^{-1} L$ at STP. [nearest integer]
[Given : Assume $CO$ as ideal gas, Mass of $C$ is $12\, g \,mol ^{-1}$, Mass of $O$ is $16 \,g \,mol ^{-1}$ and molar volume of an ideal gas at STP is $22.7\, L\, mol ^{-1}$ ]

Hint:

At STP, one mole of an ideal gas occupies 22.7 L. Use this fact for calculations involving gases at STP.

Answer 1

Correct Answer: 227

To solve this problem, we need to calculate the volume of carbon monoxide (CO) produced when 12 g of carbon is burnt in 48 g of oxygen at STP.

The reaction equation is: 2 C_(s) + O_2(g) → 2 CO(g)

First, calculate the moles of carbon and oxygen:

• Molar mass of C = 12 g/mol, so moles of C = 12 g / 12 g/mol = 1 mole
• Molar mass of O₂ = 2 × 16 g/mol = 32 g/mol, so moles of O₂ = 48 g / 32 g/mol = 1.5 moles

 

The stoichiometry of the reaction indicates that 2 moles of C react with 1 mole of O₂ to produce 2 moles of CO.

Here, carbon is the limiting reagent since 1 mole C requires 0.5 moles of O₂, and we have more oxygen than required (1.5 moles).

Thus, 1 mole of C will yield 1 mole of CO.

The molar volume of an ideal gas at STP is 22.7 L/mol. Therefore, the volume of CO produced is:
Volume of CO = 1 mole × 22.7 L/mol = 22.7 L

Convert this volume to the given scale:
22.7 L = 2.27 × 10¹ L

Rounded to the nearest integer: Volume = 23 × 10^-1 L