# 1.91 g H2 is allowed to react with 9.86 g N2, producing 1.74 g NH3. What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

Feb 16, 2018

$\text{Theoretical yield} = 32.5 g N {H}_{3}$

#### Explanation:

1. Write and balance the equation

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

2. Find the molar masses of the reactants and the product. Values are obtainable from the periodic table.

${H}_{2} = 2 \text{g/mol}$
${N}_{2} = 28 \text{g/mol}$
$N {H}_{3} = 17 \text{g/mol}$

3. Given the masses of the reactants, the number of moles can be calculated as illustrated below.

$\eta {H}_{2} = 1.91 \cancel{g {H}_{2}} \times \frac{1 m o l {H}_{2}}{2 \cancel{g {H}_{2}}} = 0.955 m o l {H}_{2}$
$\eta {N}_{2} = 9.86 \cancel{g {N}_{2}} \times \frac{1 m o l {N}_{2}}{28 \cancel{g {N}_{2}}} = 0.352 m o l {N}_{2}$

4. Now, identifying how much each reactant requires if the reaction goes into completion. Refer to the balanced equation for the needed molar ratio.
a. $\underline{\eta {H}_{2} = 0.955 m o l \text{ available}} :$
$\eta {N}_{2} = 0.955 \cancel{m o l {H}_{2}} \times \frac{1 m o l {N}_{2}}{3 \cancel{m o l {H}_{2}}} = 0.318 m o l {N}_{2}$
Thus;
$0.955 m o l {H}_{2} \equiv 0.318 m o l {N}_{2}$
b. ul(etaN_2=0.352mol " available":
$\eta {H}_{2} = 0.352 \cancel{m o l {N}_{2}} \times \frac{3 m o l {H}_{2}}{1 \cancel{m o l {N}_{2}}} = 1.056 m o l {H}_{2}$
Thus;
$0.352 m o l {N}_{2} \equiv 1.056 m o l {H}_{2}$
5. Find the limiting reactant through comparison: the required and the available number of moles; i.e.,
$\frac{{H}_{2} \text{ available")/(0.955molH_2)<(H_2" required}}{1.056 m o l {H}_{2}}$
color(red)(H_2 " is the limiting reactant."
$\frac{{N}_{2} \text{ available")/(0.352molN_2)>(N_2" required}}{0.318 m o l {N}_{2}}$
${N}_{2} \text{ is the excess reactant.}$

6. Knowing the limiting reactant, the theoretical yield $\left(T Y\right)$ can be computed as shown below. Refer to the balanced equation for the mole ratio and the molar mass of the $N {H}_{3}$ as reflected in $\text{step} 2$ for the following series of conversions:
$= 0.955 \cancel{m o l {H}_{2}} \times \frac{2 m o l N {H}_{3}}{1 \cancel{m o l {H}_{2}}}$
$= 1.91 \cancel{m o l N {H}_{3}} \times \frac{17 g N {H}_{3}}{1 \cancel{m o l N {H}_{3}}}$
$= 32.5 g N {H}_{3}$