Đáp án:
\(\begin{array}{l}
b)\\
\% C{H_4} = 21,62\% \\
\% {C_4}{H_{10}} = 78,38\% \\
c)\\
{V_{{O_2}}} = 19,04l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \frac{{11,2}}{{22,4}} = 0,5mol\\
hh:C{H_4}(a\,mol),{C_4}{H_{10}}(b\,mol)\\
16a + 58b = 7,4(1)\\
a + 4b = 0,5(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,1\\
{m_{C{H_4}}} = n \times M = 0,1 \times 16 = 1,6g\\
{m_{{C_4}{H_{10}}}} = 7,4 - 1,6 = 5,8g\\
\% C{H_4} = \dfrac{{1,6}}{{7,4}} \times 100\% = 21,62\% \\
\% {C_4}{H_{10}} = 100 - 21,62 = 78,38\% \\
c)\\
{n_{{O_2}}} = 2{n_{C{H_4}}} + \dfrac{{13}}{2}{n_{{C_4}{H_{10}}}} = 0,85mol\\
{V_{{O_2}}} = n \times 22,4 = 0,85 \times 22,4 = 19,04l\\
\end{array}\)
