Đáp án:
a)
\(\begin{array}{l}
\% {V_{C{l_2}}} = 44,44\% \\
\% {V_{{H_2}}} = 55,56\% \\
b)\\
\% {V_{C{l_2}}} = 25,56\% \\
\% {V_{{H_2}}} = 33,67\% \\
\% {V_{HCl}} = 37,77\% \\
c)\\
H = 42,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
hh:C{l_2}(a\,mol),{H_2}(b\,mol)\\
{n_{hh}} = \dfrac{{2,016}}{{22,4}} = 0,09\,mol\\
{M_{hh}} = 8,1667 \times 4 = 32,6668g/mol\\
{m_{hh}} = 32,6668 \times 0,09 = 2,94g\\
\left\{ \begin{array}{l}
a + b = 0,09\\
71a + 2b = 2,94
\end{array} \right.\\
\Rightarrow a = 0,04;b = 0,05\\
\% {V_{C{l_2}}} = \dfrac{{0,04}}{{0,09}} \times 100\% = 44,44\% \\
\% {V_{{H_2}}} = 100 - 44,44 = 55,56\% \\
b)\\
{H_2} + C{l_2} \to 2HCl(1)\\
C{l_2} + {H_2}O \to HCl + HClO(2)\\
HCl + AgN{O_3} \to AgCl + HN{O_3}\\
{n_{AgCl}} = \dfrac{{8,16}}{{143,5}} \approx 0,057\,mol\\
\text{ Gọi a là số mol $Cl_2$ tham gia phản ứng }\\
{n_{HCl(1)}} = 2{n_{C{l_2}}} \text{ tham gia }= 2a\,mol\\
{n_{HCl(2)}} = {n_{C{l_2}}} \text{ dư}= 0,04 - a\,mol\\
\Rightarrow 2a + 0,04 - a = 0,057 \Leftrightarrow a = 0,017\,mol\\
{n_{{H_2}}} = 0,05 - 0,017 = 0,033\,mol\\
{n_{C{l_2}}} = 0,04 - 0,017 = 0,023\,mol\\
\% {V_{C{l_2}}} = \dfrac{{0,023}}{{0,023 + 0,033 + 0,034}} \times 100\% = 25,56\% \\
\% {V_{{H_2}}} = \dfrac{{0,033}}{{0,023 + 0,033 + 0,034}} \times 100\% = 33,67\% \\
\% {V_{HCl}} = 100 - 36,67 - 25,56 = 37,77\% \\
c)\\
H = \dfrac{{0,017}}{{0,04}} \times 100\% = 42,5\%
\end{array}\)
