Em tham khảo nha:

\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_6} + 7{O_2} \xrightarrow{t^0} 4C{O_2} + 6{H_2}O\\
hh:C{H_4}(a\,mol),{C_2}{H_6}(b\,mol)\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{O_2}}} = \dfrac{{24}}{{32}} = 0,75\,mol\\
\left\{ \begin{array}{l}
a + b = 0,3\\
2a + 3,5b = 0,75
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{C{H_4}}} = 0,2 \times 16 = 3,2g\\
{m_{{C_2}{H_6}}} = 0,1 \times 30 = 3g\\
b)\\
{n_{C{O_2}}} = 0,2 + 0,1 \times 2 = 0,4\,mol\\
{V_{C{O_2}}} = 0,4 \times 22,4 = 8,96l
\end{array}\)

 

Đáp án:

Giải thích các bước giải:

$a)$  

$n_{hh}=\dfrac{6,72}{22,4}=0,3(mol)$

$n_{O_2}=\dfrac{24}{32}=0,75(mol)$

Phản ứng xảy ra:

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

$2C_2H_6+7O_2\xrightarrow{t^o}4CO_2+6H_2O$

Đặt: $n_{CH_4}=x(mol), n_{C_2H_6}=y(mol)$

$\to x+y=0,3(mol)$

$n_{O_2}=2x+3,5y=0,75(mol)$

$\to x=0,2(mol), y=0,1(mol)$

$m_{CH_4}=0,2.16=3,2(g)$

$m_{C_2H_6}=0,1.30=3(g)$

$b)$ Theo phương trình:

$n_{CO_2}=n_{CH_4}+2n_{C_2H_6}=0,4(mol)$

$V_{CO_2}=0,4.22,4=8,96(l)$