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

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Some Basic Concepts of Chemistry

Assume carbon burns according to following equation :

2 C_{(s)} + O_{2 (g)} \to 2 CO (g)

When 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is ____

\times 1 0^{- 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}

]

Answer: 227

Solution

The amounts taken are

12/12 = 1 mol

of carbon and

48/32 = 1.5 mol

of oxygen. By the equation

2 C + O_{2} \to 2 CO

, one mole of carbon needs only half a mole of oxygen, so oxygen is in excess and carbon is the limiting reactant, yielding one mole of carbon monoxide. At STP this occupies

1 \times 22.7 = 22.7 L

, which is

227 \times 1 0^{- 1} L

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Some Basic Concepts of Chemistry

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