Question

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the reaction proceeds to 82.0% yield?
2N2​(g)+O2​(g)= 2N2​O(g)

Answer 1

Answer;

=259 ml

Explanation;

-According to Gay Lussac's Law of Combining Volumes when gases react, they do so in volumes which have a simple ratio to one another, and to the volume of the product formed if gaseous, provided the temperature and pressure remain constant.

-Thus; from the volume of nitrogen and oxygen gases; we have; 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas.

-Therefore, nitrogen gas is the limiting reactant, and for each mole of nitrogen gas used, we will get 1 mole of N2O. This means the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, thus, 100% yield will produce 316 mL.

However, with 82% yield the volume would be; 316 × 82/100 =259 ml

Therefore; the volume of N2O at 82% yield will be 259 ml

Answer 2

259 mL Assuming the the nitrogen and oxygen gasses are at the same pressures, then the ratios of the moles of the gasses will be in proportion to their volumes according to the ideal gas law. So we have 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas. This means that the nitrogen gas will be the limiting reactant. For each mole of nitrogen gas used, we will get 1 mole of N2O. So the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, so 100% yield will produce 316 mL. But since we only have 82.0% yield, that means that the volume of N2O produced will be 316 mL * 0.820 = 259.12 mL. Rounding to 3 significant figures, gives 259 mL. So assuming constant pressure and temperature, the volume of N2O will be 259 mL.