Step 1

$\mathbf{5}\mathbf{.}\mathbf{60}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}{\mathbf{17}\mathbf{.}\mathbf{031}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}$** = 0.3288 mol NH _{4}Cl**

**$\mathbf{4}\mathbf{.}\mathbf{60}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{HCl}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}}}{\mathbf{36}\mathbf{.}\mathbf{46}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{HCl}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}\mathbf{Cl}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}}}$ = 0.1262 mol NH _{4}Cl**

HCl → limiting reactant

Ammonia, NH_{3}(g), and hydrogen chloride, HCl(g), react to form solid ammonium chloride, NH_{4}Cl(s):

NH_{3}(g) + HCl(g) → NH_{4}Cl(s)

Two 2.50 L flasks at 30.0 ^{o}C are connected by a stopcock, as shown in the drawing

One flask contains 5.60g NH_{3}(g), and the other contains 4.60 g HCl(g). When the stopcock is opened, the gases react until one is completely consumed.

What mass of ammonium chloride will be formed?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

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