Using the equation $N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$ , how many grams of hydrogen must react if the reaction needs 127 grams of $N H_{3}$ $NH_3$ ?

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Using the equation $N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$ , how many grams of hydrogen must react if the reaction needs 127 grams of $N H_{3}$ $NH_3$ ?

3 Answers

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Answer 1 · 2016-07-02T14:49:41

22.6 g $H_{2}$ $"H"_2$

Explanation:

${3H}_{2} : : {2NH}_{3}$ $"3H"_2::"2NH"_3$

$\frac{3}{2} H_{2} : : {1NH}_{3}$ $3/2"H"_2::"1NH"_3$

${Moles of NH}_{3} = \frac{127}{17.0}$ $"Moles of NH"_3=127/17.0$

So ${moles of H}_{2} = \frac{3}{2} \cdot \frac{127}{17.0} = 11.2$ $"moles of H"_2=3/2*127/17.0 = 11.2$

Mass of $H_{2} = 11.2 mol \times 2.016 g/mol = 22.6 g$ $"H"_2 = "11.2 mol" × "2.016 g/mol" = "22.6 g"$

Answer 2 · 2016-07-02T15:11:25

${22.6 g of H}_{2}$ $"22.6 g of H"_2$ must react.

Given: Mass of ${NH}_{3}$ $"NH"_3$ and the chemical equation.

Find: Mass of $H_{2}$ $"H"_2$ .

Strategy:

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

(a) We start with the balanced chemical equation for the reaction.

(b) We can use the molar mass of ${NH}_{3}$ $"NH"_3$ to find the moles of ${NH}_{3}$ $"NH"_3$ .

(c) We can use the molar ratio from the equation to convert moles of ${NH}_{3}$ $"NH"_3$ to moles of $H_{2}$ $"H"_2$ .

${moles of NH}_{3} molar ratio X l - ------ \to {moles of H}_{2}$ $"moles of NH"_3 stackrelcolor(blue)("molar ratio"color(white)(Xl))( →) "moles of H"_2$

(d) Then we can use the molar mass to convert the moles of $H_{2}$ $"H"_2$ to mass of $H_{2}$ $"H"_2$ .

Our complete strategy is:

${Mass of NH}_{3} molar mass X l - ------ \to {moles of NH}_{3} molar ratio X l - ------ \to {moles of H}_{2} molar mass X l - ------ \to {mass of H}_{2}$ $"Mass of NH"_3stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of NH"_3stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of H"_2 stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "mass of H"_2$

Solution

(a) The balanced equation is

$N_{2} + {3H}_{2} \to {2NH}_{3}$ $"N"_2 + "3H"_2 → "2NH"_3$

(b) Calculate moles of ${NH}_{3}$ $"NH"_3$

$127 {g NH}_{3} \times \frac{{1 mol NH}_{3}}{17.03 {g NH}_{3}} = {7.457 mol NH}_{3}$ $127 color(red)(cancel(color(black)("g NH"_3))) × ("1 mol NH"_3)/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "7.457 mol NH"_3$

(c) Calculate moles of $H_{2}$ $"H"_2$

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

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

(d) Calculate the mass of $H_{2}$ $"H"_2$

$11.19 {mol H}_{2} \times \frac{{2.016 g H}_{2}}{1 {mol H}_{2}} = {22.6 g H}_{2}$ $11.19 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "22.6 g H"_2$

**Answer: ** ${22.6 g of H}_{2}$ $"22.6 g of H"_2$ must react.

Answer 3 · 2016-07-02T15:17:54

$22.58 g H_{2}$ $22.58 g H_2$

Explanation:

$N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$

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

$127 g N H_{3} \cdot \frac{1 m o l N H_{3}}{17.04 g N H_{3}} \cdot \frac{3 m o l H_{2}}{2 m o l N H_{3}} \cdot \frac{2.02 g H_{2}}{1 m o l H_{2}} =$ $127g NH_3 * (1 mol NH_3)/(17.04 g NH_3)*(3 mol H_2)/(2 mol NH_3)*(2.02 g H_2)/(1 mol H_2) =$

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

$14.01 + 3.03 = 17.04 g$ $14.01+3.03=17.04g$

$127 g N H_{3} \cdot \frac{1 m o l N H_{3}}{17.04 g N H_{3}} \cdot \frac{3 m o l H_{2}}{2 m o l N H_{3}} \cdot \frac{2.02 g H_{2}}{1 m o l H_{2}} =$ $127cancel(g NH_3) * (1 cancel(mol NH_3))/(17.04 cancel(g NH_3))*(3 cancel(mol H_2))/(2 cancel(mol NH_3))*(2.02 g H_2)/(1 cancel(mol H_2)) =$

$= 22.58 g H_{2}$ $=22.58 g H_2$

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Using the equation $N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$ , how many grams of hydrogen must react if the reaction needs 127 grams of $N H_{3}$ $NH_3$ ?

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Using the equation N2+3H2→2NH3N_2 + 3H_2 -> 2NH_3, how many grams of hydrogen must react if the reaction needs 127 grams of NH3NH_3?

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Explanation:

Explanation:

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Using the equation $N_{2} + 3 H_{2} \to 2 N H_{3}$ $N_2 + 3H_2 -> 2NH_3$ , how many grams of hydrogen must react if the reaction needs 127 grams of $N H_{3}$ $NH_3$ ?