Đáp án:
B
Giải thích các bước giải:
Phản ứng xảy ra:
\(2CO + {O_2}\xrightarrow{{}}2C{O_2}\)
\(2{H_2} + {O_2}\xrightarrow{{}}2{H_2}O\)
Ta có: \({n_{C{O_2}}} = {n_{CO}} = \frac{{13,2}}{{44}} = 0,3{\text{ mol}}\)
\({n_{{O_2}}} = \frac{{12,8}}{{16.2}} = 0,4{\text{ mol = }}\frac{1}{2}{n_{CO}} + \frac{1}{2}{n_{{H_2}}} \to {n_{{H_2}}} = 0,5{\text{ mol}}\)
\(\to {m_{CO}} = 0,3.(12 + 16) = 8,4{\text{ gam; }}{{\text{m}}_{{H_2}}} = 0,5.2 = 1{\text{ gam}}\)
\(\to \% {m_{CO}} = \frac{{8,4}}{{8,4 + 1}} = 89,36\% \to \% {m_{{H_2}}} = 10,64\% \)