# If at a certain temperature, "6.8 g" of "NH"_3 completely decomposes to form "H"_2(g) and "N"_2(g), how many mols of "Al"(s) would be needed to react with excess aqueous "HCl" to form the same amount of "H"_2(g)?

Mar 16, 2017

The idea is to see how many mols of ${\text{H}}_{2}$ you would have made in one reaction and reuse that value in the context of the other reaction.

Assuming you mean $\text{6.8 g}$ of ${\text{NH}}_{3}$ decomposes as:

$2 {\text{NH"_3(g) rightleftharpoons 3"H"_2(g) + "N}}_{2} \left(g\right)$

and assuming the reaction goes to completion, we make:

"6.8 g NH"_3 xx ("1 mol NH"_3)/(14.007 + 3xx1.0079 "g NH"_3) xx ("3 mols H"_2)/("2 mol NH"_3)

$= {\text{0.5989 mols H}}_{2}$

To make that many mols of ${\text{H}}_{2}$ in the reaction of

${\text{Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + "AlCl}}_{3} \left(a q\right)$

You'll need

"0.5989 mols H"_2(g) xx ("1 mol Al"(s))/("3 mols H"_2(g))

$=$ $\text{0.1996 mols Al}$

To 2 sig figs, it would be $\textcolor{b l u e}{\text{0.20 mols Al}}$.

The forward ammonia reaction was called the Haber process, i.e. the production of ${\text{NH}}_{3} \left(g\right)$ from ${\text{N}}_{2} \left(g\right)$ and ${\text{H}}_{2} \left(g\right)$ at high pressures and temperatures. We looked at the reverse reaction.

The aluminum reaction is a common example of placing a metal into water and reacting it with a strong acid to produce a gas. Since your strong acid was $\text{HCl}$, you made ${\text{H}}_{2}$ gas.

(If you had used ${\text{HNO}}_{3}$, you might have made the brown ${\text{NO}}_{2}$ gas.)