# Question f171f

$30 g r a m s$
The molecular weight of $S i {O}_{2}$ is 60g/mol, so 50.0g is 0.833 moles and there are therefore 0.833 moles of ${O}_{2}$ available for reaction.
Whatever the reaction is, assuming that all of the oxygen is reacted to form water, and oxygen atoms combine in a 1:2 ratio with hydrogen atoms 2H_2 + O_2 → 2H_2O# we can see that each mole of ${O}_{2}$ can produce two moles of water. Therefore, the moles of water produced is $2 \cdot 0.833 = 1.67$. The molecular weight of water is 18, so this number of moles will yield $1.67 m o l \cdot 18 \frac{g}{\text{mol}} = 30 g$ of water.