# How many moles of water are produced when "2.15 moles" of oxygen gas react with excess hydrogen gas?

$\text{4.30 moles}$
$\text{I mol}$ of ${\text{O}}_{2}$ produces $\text{2 moles}$ of $\text{H"_2"O}$, so $\text{2.15 moles}$ of ${\text{O}}_{2}$ produce
2/1×2.15 = "4.30 moles H"_2"O"