Đáp án:

 a=0,2 gam

Giải thích các bước giải:

 Ta có: \({m_{hh}} = {m_{S{O_2}}} + {m_{C{l_2}}} + {m_{{H_2}}} = 12,8 + 7,1 + a = 19,9 + a{\text{ gam}}\)

\({n_{S{O_2}}} = \frac{{12,8}}{{32 + 16.2}} = 0,2{\text{ mol; }}{{\text{n}}_{C{l_2}}} = \frac{{7,1}}{{35,5.2}} = 0,1{\text{ mol; }}{{\text{n}}_{{H_2}}} = \frac{a}{2} = 0,5a{\text{ mol}} \to {{\text{n}}_{hh}} = {n_{S{O_2}}} + {n_{C{l_2}}} + {n_{{H_2}}} = 0,2 + 0,1 + 0,5a = 0,3 + 0,5a{\text{ mol}}\)

\({M_{hh}} = 20,1{M_{{H_2}}} = 20,1.2 = 40,2 = \frac{{{m_{hh}}}}{{{n_{hh}}}} = \frac{{19,9 + a}}{{0,3 + a}} \to a = 0,2{\text{ gam}}\)

Ta có: 

\(\% {m_{S{O_2}}} = \frac{{12,8}}{{20,1}} = 63,68\% ;\% {m_{C{l_2}}} = \frac{{7,1}}{{20,1}} = 35,32\%  \to \% {m_{{H_2}}} = 1\% \)