Question

The combustion of carbon monoxide yields carbon dioxide. The volume of oxygen gas needed to produce 22 g of carbon dioxide at STP is

• A. 4.0 L
• B. 5.6 L
• C. 11 L
• D. 22 L

Hint:

At STP, 1 mole of gas occupies 22.4 L, and this relationship can be used to calculate the volume of gases in chemical reactions.

Answer 1

The Correct Option is B

The balanced chemical equation for the combustion of carbon monoxide is: \[ 2 \, \text{CO} + \text{O}_2 \to 2 \, \text{CO}_2 \] From the equation, 2 moles of CO react with 1 mole of O\(_2\) to produce 2 moles of CO\(_2\).
The molar mass of CO is 28 g/mol, and the molar mass of CO\(_2\) is 44 g/mol.
To calculate the volume of oxygen needed to produce 22 g of CO\(_2\), we first find the number of moles of CO\(_2\): \[ \text{moles of CO}_2 = \frac{22 \, \text{g}}{44 \, \text{g/mol}} = 0.5 \, \text{mol} \] Since 1 mole of O\(_2\) reacts with 2 moles of CO to produce 2 moles of CO\(_2\), the moles of O\(_2\) required are: \[ \text{moles of O}_2 = 0.5 \, \text{mol} \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol CO}_2} = 0.25 \, \text{mol O}_2 \] At STP, 1 mole of any ideal gas occupies 22.4 L, so the volume of O\(_2\) is: \[ \text{volume of O}_2 = 0.25 \, \text{mol} \times 22.4 \, \text{L/mol} = 5.6 \, \text{L} \]
Thus, the correct answer is (b).