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

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

Step-by-step 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) * \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) * \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) * \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) * \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) * \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) * \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)`.