Đáp án:
%V CH4=33,33%; %V H2S=66,67%
V O2=11200 ml
Giải thích các bước giải:
Ta có: \({V_A} = 6720{\text{ ml }}\)
Vì \({V_{C{H_4}}} + {V_{{H_2}S}} = {V_A} = 6,72;{\text{ }}{{\text{V}}_{C{H_4}}}:{V_{{H_2}S}} = 1:2 \to {V_{C{H_4}}} = 2240{\text{ ml; }}{{\text{V}}_{{H_2}S}} = 4480{\text{ml}}\)
Ta có: \(\% {V_{C{H_4}}} = \frac{{2240}}{{6720}} = 33,33\% \to \% {V_{{H_2}S}} = 66,67\% \)
Phản ứng cháy:
\(C{H_4} + 2{O_2}\xrightarrow{{}}C{O_2} + 2{H_2}O\)
\(2{H_2}S + 3{O_2}\xrightarrow{{}}2S{O_2} + 2{H_2}O\)
\(\to {V_{{O_2}}} = 2{V_{C{H_4}}} + \frac{3}{2}{V_{{H_2}S}} = 2240.2 + 4480.\frac{3}{2} = 11200{\text{ ml}}\)