Đáp án:

a) $\% {V_{{H_2}}} = 55,55\% ;\\\% {V_{C{l_2}}} = 44,45\% $

b) $\% {V_{HCl}} = 63,33\% ;\\\% {V_{C{l_2}}} = 12,78\% ;\\\% {V_{{H_2}}} = 23,89\% $

c) $H = 71,25\% $

Giải thích các bước giải:

 a)

${M_A} = 8,1667.4 = 32,6668$;

${n_A} = \dfrac{{2,016}}{{22,4}} = 0,09mol$

Phương pháp đường chéo: 

$\begin{array}{*{20}{c}} {{H_2}:2}&{}&{}&{}&{38,3332} \\ {}& \searrow &{}& \nearrow &{} \\ {}&{}&{32,6668}&{}&{} \\ {}& \nearrow &{}& \searrow &{} \\ {C{l_2}:71}&{}&{}&{}&{30,6668} \end{array}$

$ \Rightarrow \dfrac{{{V_{{H_2}}}}}{{{V_{C{l_2}}}}} = \dfrac{5}{4}$

$\begin{gathered} \Rightarrow \% {V_{{H_2}}} = \dfrac{5}{9}.100\% = 55,55\% \hfill \\ \% {V_{C{l_2}}} = 100 - 55,55 = 44,45\% \hfill \\ \end{gathered} $

b)

${n_{AgCl}} = \dfrac{{8,16}}{{143,5}} = 0,057mol$

${n_{{H_2}}} = 0,05mol;{n_{C{l_2}}} = 0,05mol$

$C{l_2} + {H_2} \to 2HCl$

 $0,0285$   $0,0285$ $←$  $0,057$

$AgN{O_3} + HCl \to AgCl + HN{O_3}$

                   $0,057$ $←$  $0,057$

⇒ Hỗn hợp B gồm $HCl(0,057{\text{ mol); C}}{{\text{l}}_2}(0,0115{\text{ mol); }}{{\text{H}}_2}(0,0215{\text{ mol)}}$

$ \Rightarrow {n_B} = 0,057 + 0,0115 + 0,0215 = 0,09mol$

$\begin{gathered} \Rightarrow \% {V_{HCl}} = \dfrac{{0,057}}{{0,09}}.100\% = 63,33\% \hfill \\ \% {V_{C{l_2}}} = \dfrac{{0,0115}}{{0,09}}.100\% = 12,78\% \hfill \\ \% {V_{{H_2}}} = 100 - 63,33 - 12,78 = 23,89\% \hfill \\ \end{gathered} $

c)

Do ${n_{{H_2}}} > {n_{C{l_2}}}$ ⇒ $C{l_2}$ hết trước $H{_2}$

⇒ Hiệu suất tính theo $C{l_2}$

$ \Rightarrow H = \dfrac{{0,0285}}{{0,04}}.100\%  = 71,25\% $