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\% $
