# How many grams of N_2 are required to completely react with 3.03 grams of H_2 for the equation N_2 + 3H_2 -> 2NH_3?

Dec 10, 2015

$7.01 \text{g}$

#### Explanation:

Notice the coefficients of ${N}_{2}$ and ${H}_{2}$.

Since the equation has an ${N}_{2}$ and $3 {H}_{2}$, $3$ times as many moles of ${H}_{2}$ will be used as ${N}_{2}$ assuming a complete reaction.

Since we're given the mass of ${H}_{2}$ being used, we can divide that by the molar mass of ${H}_{2}$ to determine how many moles of ${H}_{2}$ will be consumed. We can divide that mole amount by $3$ to find how many moles of ${N}_{2}$ the reaction will use, and multiply that by the molar mass of ${N}_{2}$ to find how many grams we should use.

$3.03 \text{g"H_2xx(1"mol"H_2)/(2.02"g"H_2)xx(1"mol"N_2)/(3"mol"H_2)xx(14.02"g"N_2)/(1"mol"N_2)=7.01"g} {N}_{2}$