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?

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