# What is the number of moles of O_2 that will react with 10.0g of H_2 to form water in the equation 2H_2 + O_2 -> 2H_2O?

May 29, 2017

2.48 moles of ${O}_{2}$ will react with 10.0g of ${H}_{2}$ to form water.

#### Explanation:

For this problem, we should implement the process of stoichiometry. First, convert the grams of ${H}_{2}$ into moles:

$10.0 g {H}_{2} \times \frac{1 m o l {H}_{2}}{2.01588 g {H}_{2}}$

Then multiply by your mole to mole ratio:

$10.0 g {H}_{2} \times \frac{1 m o l {H}_{2}}{2.01588 g {H}_{2}} \times \frac{1 m o l {O}_{2}}{2 m o l {H}_{2}}$

Cancel out common terms and perform the math to get your answer:

$10.0 \cancel{g {H}_{2}} \times \frac{1 \cancel{m o l {H}_{2}}}{2.01588 \cancel{g {H}_{2}}} \times \frac{1 m o l {O}_{2}}{2 \cancel{m o l {H}_{2}}} = 2.48 m o l {O}_{2}$

The answer has three sig figs because of the given 10.0g.