Using the equation N_2 + 3H_2 -> 2NH_3, how many grams of hydrogen must react if the reaction needs 127 grams of NH_3?

Jul 2, 2016

22.6 g ${\text{H}}_{2}$

Explanation:

${\text{3H"_2::"2NH}}_{3}$

$\frac{3}{2} {\text{H"_2::"1NH}}_{3}$

${\text{Moles of NH}}_{3} = \frac{127}{17.0}$

So ${\text{moles of H}}_{2} = \frac{3}{2} \cdot \frac{127}{17.0} = 11.2$

Mass of $\text{H"_2 = "11.2 mol" × "2.016 g/mol" = "22.6 g}$

Jul 2, 2016

${\text{22.6 g of H}}_{2}$ must react.

Given: Mass of ${\text{NH}}_{3}$ and the chemical equation.

Find: Mass of ${\text{H}}_{2}$.

Strategy:

The central part of any stoichiometry problem is to convert moles of something to moles of something else.

(b) We can use the molar mass of ${\text{NH}}_{3}$ to find the moles of ${\text{NH}}_{3}$.

(c) We can use the molar ratio from the equation to convert moles of ${\text{NH}}_{3}$ to moles of ${\text{H}}_{2}$.

${\text{moles of NH"_3 stackrelcolor(blue)("molar ratio"color(white)(Xl))( →) "moles of H}}_{2}$

(d) Then we can use the molar mass to convert the moles of ${\text{H}}_{2}$ to mass of ${\text{H}}_{2}$.

Our complete strategy is:

${\text{Mass of NH"_3stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of NH"_3stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of H"_2 stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "mass of H}}_{2}$

Solution

(a) The balanced equation is

${\text{N"_2 + "3H"_2 → "2NH}}_{3}$

(b) Calculate moles of ${\text{NH}}_{3}$

127 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "7.457 mol NH"_3

(c) Calculate moles of ${\text{H}}_{2}$

The molar ratio of ${\text{H}}_{2}$ to ${\text{NH}}_{3}$ is ("3 mol H"_2)/("2 mol NH"_3)".

${\text{Moles of H"_2 = 7.457 color(red)(cancel(color(black)("mol NH"_3))) × ("3 mol H"_2)/(2 color(red)(cancel(color(black)("mol NH"_3)))) = "11.19 mol H}}_{2}$

(d) Calculate the mass of ${\text{H}}_{2}$

11.19 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "22.6 g H"_2

Answer: ${\text{22.6 g of H}}_{2}$ must react.

Jul 2, 2016

$22.58 g {H}_{2}$

Explanation:

${N}_{2} + 3 {H}_{2} \to 2 N {H}_{3}$

If $127$ grams of necessary we can use stoichiometry conversions to find the mass of Hydrogen required.

$127 g N {H}_{3} \cdot \frac{1 m o l N {H}_{3}}{17.04 g N {H}_{3}} \cdot \frac{3 m o l {H}_{2}}{2 m o l N {H}_{3}} \cdot \frac{2.02 g {H}_{2}}{1 m o l {H}_{2}} =$

$N = 1 \cdot 14.01 g = 14.01$
$H = 3 \cdot 1.01 g = 3.03$

$14.01 + 3.03 = 17.04 g$

$127 \cancel{g N {H}_{3}} \cdot \frac{1 \cancel{m o l N {H}_{3}}}{17.04 \cancel{g N {H}_{3}}} \cdot \frac{3 \cancel{m o l {H}_{2}}}{2 \cancel{m o l N {H}_{3}}} \cdot \frac{2.02 g {H}_{2}}{1 \cancel{m o l {H}_{2}}} =$

$= 22.58 g {H}_{2}$