# How many molecules of NH_3 are produced from 4.82*10^-4 g of H_2?

Feb 27, 2016

$9.58 \times {10}^{19} \text{molecules} N {H}_{3}$

#### Explanation:

The reaction of making $N {H}_{3}$ from ${H}_{2}$ is the following:

$3 {H}_{2} \left(g\right) + {N}_{2} \left(g\right) \to 2 N {H}_{3} \left(g\right)$

Therefore, using dimensional analysis, we can use the molar ratio from the balanced equation to find the number of molecules of $N {H}_{3}$ produced by:

?"molecules"NH_3=4.82xx10^(-4)cancel(gH_2)xx(1cancel(molH_2))/(2.02cancel(gH_2))xx(2cancel(molNH_3))/(3cancel(molH_2))xx(6.022xx10^(23)"molecules"NH_3)/(1cancel(molNH_3))=9.58xx10^(19)"molecules"NH_3