# How many grams of O_2(g) are needed to completely burn 21.4 g of C_3H_8(g) in the reaction C_3H_8(g) + 5O_2(g) -> 3CO_2(g) + 4H_2O(g)?

Find the number of moles of ${C}_{3} {H}_{8}$: $n = \frac{m}{M} = \frac{21.4}{44} = 0.49$ $m o l$. Each mole of ${C}_{3} {H}_{8}$ requires $5$ $m o l$ of ${O}_{2}$, so we need $5 \times 0.49 = 2.45$ $m o l$. The mass will be $m = n M = 2.45 \times 32 = 78.4$ $g$.