Đáp án:

\(\begin{array}{l}
a)\\
hh:{C_3}{H_8},{C_4}{H_{10}}\\
b)\\
{V_{kk}} = 18,48l\\
c)\\
\% {m_{{C_3}{H_8}}} = 60,27\% \\
\% {m_{{C_4}{H_{10}}}} = 39,73\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{10}}{{100}} = 0,1\,mol\\
m = {m_{C{O_2}}} + {m_{{H_2}O}}\\
 \Leftrightarrow 0,1 \times 44 + {n_{{H_2}O}} \times 18 = 6,74\\
 \Leftrightarrow {n_{{H_2}O}} = 0,13\,mol\\
{n_{C{O_2}}} < {n_{{H_2}O}} \Rightarrow hh\,ankan({C_{\overline n }}{H_{2\overline n  + 2}})\\
{n_{ankan}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,13 - 0,1 = 0,03\,mol\\
\overline n  = \dfrac{{0,1}}{{0,03}} = 3,33 \Rightarrow hh:{C_3}{H_8},{C_4}{H_{10}}\\
b)\\
{C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O\\
2{C_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O\\
hh:{C_3}{H_8}(a\,mol),{C_4}{H_{10}}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,03\\
3a + 4b = 0,1
\end{array} \right.\\
 \Rightarrow a = 0,02;b = 0,01\\
{n_{{O_2}}} = 0,02 \times 5 + 0,01 \times \frac{{13}}{2} = 0,165\,mol\\
{V_{kk}} = 0,165 \times 5 \times 22,4 = 18,48l\\
c)\\
\% {m_{{C_3}{H_8}}} = \dfrac{{0,02 \times 44}}{{0,02 \times 44 + 0,01 \times 58}} \times 100\%  = 60,27\% \\
\% {m_{{C_4}{H_{10}}}} = 100 - 60,27 = 39,73\% 
\end{array}\)