Question

Acetylene reacts with oxygen to form carbon dioxide and water according to the following unbalanced reaction:

C2H2 (g) + O2 (g) → CO2 (g) + H2O (g)

If we start with 38.7 g of acetylene and 13.7 g of oxygen, how many grams of the excess reactant are leftover after the reaction is complete?

Report your answer to THREE significant figures.

Answer 1

Answer:

Mass of excess reactant leftover after reaction is complete = 34.2 g $\mathrm{C}_{2} \mathrm{H}_{2}$.

Explanation:

Acetylene reacts with oxygen to form carbon dioxide and water, equation given is unbalanced

$\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

Firstly equation is balanced:

$2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

$\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{C}_{2} \mathrm{H}_{2}$

$\text { Molar mass of } \mathrm{C}_{2} \mathrm{H}_{2} \text { is } 26.04 \mathrm{g} / \mathrm{mol}$

$\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{2}=38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{1 \mathrm{mol} c_{2} \mathrm{H}_{2}}{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}=1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}$

$1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{4 \mathrm{mol} \mathrm{Co}_{2}}{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}=2.96 \mathrm{mol} \mathrm{CO}_{2}$

$\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{O}_{2}$:

$\text { Molar mass of } \mathrm{O}_{2} \text { is } 32.00 \mathrm{g} / \mathrm{mol}$

$\text { Moles of } \mathrm{O}_{2}=13.7 \mathrm{gm} \mathrm{O}_{2} \times \frac{1 \mathrm{mol} O_{2}}{32.00 \mathrm{gm} 0_{2}}=0.428 \mathrm{mol} \mathrm{O}_{2}$

$\text { Here we come to know that } 4 \mathrm{mol} \mathrm{CO}_{2} \equiv 5 \mathrm{mol} \mathrm{O}_{2}, \text { hence }$

$0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{4 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} O_{3}}\left(4 \mathrm{mol} \mathrm{CO}_{2}\right) /\left(5 \mathrm{mol} \mathrm{O}_{2}\right)=0.3424 \mathrm{mol} \mathrm{CO}_{2}$

Here we come to know O2 is limiting reactant as it gives smaller amount of $\mathrm{CO}_{2}$ So $\mathrm{C}_{2} \mathrm{H}_{2}$ is excess reactant.

Mass of excess reactant utilized:

$0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}{5 \mathrm{mol} O_{2}}=0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}$

$0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}{1 \mathrm{mol} C_{2} \mathrm{H}_{2}}=4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}$

Mass of excess reactant leftover after reaction is complete:

$38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}-4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}=34.2 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}$.