# How many grams of water can be prepared from 10.10 grams of hydrogen gas and excess oxygen gas at standard conditions?

Jan 17, 2016

$90.9 g$

#### Explanation:

First write a balanced chemical equation for the reaction :

$2 {H}_{2} + {O}_{2} \to 2 {H}_{2} O$

Since oxygen is in excess, it implies that hydrogen is the limiting reagent and will decide how much product is formed, according to the mole ratio of the balanced equation.

${n}_{{H}_{2}} = \frac{m}{{M}_{r}} = \frac{10.10 g}{2 g / m o l} = 5.05 m o l$.

$\therefore m o \le s o f w a t e r f \mathmr{and} m e d w i l l a l s o b e$5.05mol#.

$\therefore {m}_{{H}_{2} O} = n \times {M}_{r}$

$= 5.05 \times \left(2 + 16\right)$

$= 90.9 g$.