# How many moles of oxygen would have be consumed to produce 68.1 g of water in the reaction C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O?

Approx. $20 \cdot m o l \text{ dioxygen gas}$
$\text{Moles of water}$ $=$ $\frac{68.1 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $3.78 \cdot m o l$
Given the stoichiometry, clearly there are $\frac{5}{4} \times 3.78 \cdot m o l$ $\text{dioxygen gas}$ required. What is the equivalent mass?