# How many moles of H_2 are needed to react with .80 mol of N_2?

Nov 12, 2015

$\text{2.4 mol H"_2}$ are needed to react with $\text{0.80 mol N"_2}$.

#### Explanation:

Balanced Equation

"N"_2("g")+"3H"_2("g")$\rightarrow$"2NH"_3("g")

Multiply the $\text{0.80 mol N"_2}$ times the mole ratio between $\text{H"_2}$ and $\text{N"_2}$ so that moles $\text{N"_2}$ cancel, leaving moles $\text{H"_2}$.

0.80cancel("mol N"_2)xx(3"mol H"_2)/(1cancel("mol N"_2))="2.4 mol H"_2"