Đáp án:
\(\begin{array}{l}
a)\\
\% {H_2} = 17,65\% \\
\% CO = 82,35\% \\
b)\\
{m_{{H_2}O}} = 108g\\
{m_{C{O_2}}} = 88g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
{n_{{O_2}}} = \dfrac{{89,6}}{{22,4}} = 4mol\\
hh:{H_2}(a\,mol),CO(b\,mol)\\
\left\{ \begin{array}{l}
2a + 28b = 68\\
\dfrac{1}{2}a + \dfrac{1}{2}b = 4
\end{array} \right.\\
\Rightarrow a = 6mol;b = 2mol\\
{m_{{H_2}}} = 6 \times 2 = 12g\\
\% {H_2} = \dfrac{{12}}{{68}} \times 100\% = 17,65\% \\
\% CO = 100 - 17,65 = 82,35\% \\
b)\\
{n_{{H_2}O}} = {n_{{H_2}}} = 6mol\\
{m_{{H_2}O}} = 6 \times 18 = 108g\\
{n_{C{O_2}}} = {n_{CO}} = 2mol\\
{m_{C{O_2}}} = 2 \times 44 = 88g
\end{array}\)
