What volume of ammonia, in cm^3, can be prepared by reacting 20 cm^3 of nitrogen gas with 30 cm^3 of hydrogen gas?

Assuming that all volume are measured at the same temperature and pressure and the reaction goes to completion. A. 20 B. 30 C. 40 D. 50 The answer given is A. 20, why?

Mar 8, 2016

The volume of ammonia that can be prepared is ${\text{20 cm}}^{3}$.

Explanation:

This looks like a limiting reactant problem involving volumes of gases.

That means that we must use Gay-Lussac's Law of Combining Volumes: "The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers".

Step 1. Write the balanced chemical equation.

The equation for the reaction is

$\textcolor{w h i t e}{l l} {\text{N"_2color(white)(l) +color(white)(l) "3H"_2 → "2NH}}_{3}$
$\textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m m m l} 3 \textcolor{w h i t e}{m m m l} 2$
${\text{1 cm"^3color(white)(m) "3 cm"^3color(white)(ml) "2 cm}}^{3}$

The coefficients of the balanced equation tell us that the ratio is 1:3:2.

Hence, we can say that

${\text{1 cm"^3 "of N"_2 + "3 cm"^3 "of H"_2 → "2 cm"^3 "of NH}}_{3}$

Step 2. Identify the limiting reactant.

From ${\text{N}}_{2}$: 20 color(red)(cancel(color(black)("cm"^3 "N"_2))) × ("2 cm"^3 color(white)(l)"NH"_3)/(1 color(red)(cancel(color(black)("cm"^3 "N"_2)))) = "30 cm"^3color(white)(l) "NH"_3

From ${\text{H}}_{2}$: 30 color(red)(cancel(color(black)("cm"^3 "H"_2))) × ("2 cm"^3 "NH"_3)/(3 color(red)(cancel(color(black)("cm"^3 "H"_2)))) = "20 cm"^3color(white)(l) "NH"_3

${\text{H}}_{2}$ is the limiting reactant, because it forms the smaller amount of ${\text{NH}}_{3}$.

You can't get more than ${\text{20 cm"^3color(white)(l) "NH}}_{3}$.

By the time you have formed that amount, all the ${\text{H}}_{2}$ will have been used up, and you won't be able to form any more ${\text{NH}}_{3}$.