Chapter 1, Problem 1.13NP

A mixture of oxygen and hydrogen is analyzed by passing it over hot copper oxide and through a drying tube. Hydrogen reduces the CuO according to the reaction $\text{CuO}\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\to \text{Cu}\left(\text{s}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$, and oxygen reoxidizes the copper formed according to $\text{Cu}\left(\text{s}\right)+\text{1}/{\text{2O}}_{\text{2}}\left(\text{g}\right)\to \text{CuO}\left(\text{s}\right)$. At $\text{25}°\text{C}$ and 750. Torr, $\text{172}.0{\text{ cm}}^{\text{3}}$ of the mixture yields $\text{77}.{\text{5 cm}}^{\text{3}}$ of dry oxygen measured at $\text{25}°\text{C}$ and 750. Torr after passage over CuO and the drying agent. What is the original composition of the mixture?