# How many grams of H_2 are needed to produce 11.27 g of NH_3?

Feb 23, 2016

We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.

#### Explanation:

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) \rightarrow N {H}_{3} \left(g\right)$

$11.27$ $g$ of ammonia represents $\frac{11.27 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$.

Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?

Feb 23, 2016

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.

#### Explanation:

${\text{3H"_2+"N}}_{2}$$\rightarrow$$\text{2NH"_3}$

Since the question concerns the mass of hydrogen gas needed, the assumption is that nitrogen gas is in excess.

The molar masses of $\text{H"_2}$ and $\text{NH"_3}$ are needed.

$\text{H"_2} :$ $\text{2.01588 g/mol}$
http://pubchem.ncbi.nlm.nih.gov/compound/783

$\text{NH"_3}$ $\text{17.03052 g/mol}$
http://pubchem.ncbi.nlm.nih.gov/compound/222#section=Top

We will be making the following conversions: mass $\text{NH"_3}$$\rightarrow$$\text{mol NH"_3}$$\rightarrow$$\text{mol H"_2}$$\rightarrow$$\text{mass H"_2}$

1. Determine moles of $\text{NH"_3}$ by dividing the given mass by its molar mass.

2. Determine moles of $\text{H"_2}$ by multiplying the mole ratio from the balanced equation between $\text{H"_2}$ and $\text{NH"_3}$ with moles $\text{H"_2}$ in the numerator. $\left(3 {\text{mol H"_2)/(2"mol NH}}_{3}\right)$

3. Determine mass of $\text{H"_2}$ by multiplying moles $\text{H"_2}$ by its molar mass.

$11.27 \text{g NH"_3xx(1"mol NH"_3)/(17.03052"g NH"_3)xx(3"mol H"_2)/(2"mol NH"_3)xx(2.01588"g H"_2)/(1"mol H"_2)="2.001 g H"_2}$ (rounded to four significant figures)

2.001 g hydrogen gas is needed to produce 11.27 g ammonia if nitrogen gas is present in excess.