Em tham khảo nha :

\(\begin{array}{l}
1)\\
2{C_4}{H_{10}} + 13{O_2} \xrightarrow{t^0} 8C{O_2} + 10{H_2}O\\
{n_{{C_4}{H_{10}}}} = \dfrac{{13,05}}{{58}} = 0,225mol\\
{n_{{O_2}}} = \dfrac{{13}}{2}{n_{{O_2}}} = 1,4625mol\\
{V_{{O_2}}} = 1,4625 \times 22,4 = 32,76l\\
{V_{kk}} = 5{V_{{O_2}}} = 5 \times 32,76 = 163,8g\\
2)\\
a)\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
b)\\
{n_{{O_2}}} = \dfrac{{9,6}}{{32}} = 0,3mol\\
hh:C(a\,mol),S(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
12a + 32b = 5,6
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_C} = 0,2 \times 12 = 2,4g\\
{m_S} = 5,6 - 2,4 = 3,2g\\
c)\\
\% C = \dfrac{{2,4}}{{5,6}} \times 100\%  = 42,86\% \\
\% S = 100 - 42,86 = 57,14\% \\
d)\\
{n_{C{O_2}}} = {n_C} = 0,2mol\\
{m_{C{O_2}}} = 0,2 \times 44 = 8,8g\\
{n_{S{O_2}}} = {n_S} = 0,1mol\\
{m_{S{O_2}}} = 0,1 \times 64 = 6,4g\\
\% C{O_2} = \dfrac{{0,2}}{{0,2+0,1}} \times 100\%  = 66,7\% \\
\% S{O_2} = 100 - 66,7 = 33,3\% 
\end{array}\)

Bài `1:` 

`n_{C_4H_{10}}=\frac{13,05}{58}=0,225(mol)`

Phương trình:

`2C_4H_10+13O_2\overset{t^o}{\to}8CO_2+10H_2O`

Theo phương trình, ta nhận thấy:

`n_{C_4H_{10}}.6,5=n_{O_2}`

`=> n_{O_2}=0,225.6,5=1,4625(mol)`

`=> V_{O_2}=1,4625.22,4=32,76(l)`

Do `V_{(kk)}.\frac{20}{100}=V_{O_2}`

`=> V_{O_2}.5=V_{(kk)}`

`=> V_{(kk)}=5.32,76=163,8(l)`

Bài `2:`

Cho $5,6(g) \begin{cases} C: x(mol)\\ S: y(mol)\\\end{cases}$

Ta có: `12x+32y=5,6(g)(1)`

`n_{O_2}=\frac{9,6}{32}=0,3(mol)`

`a)` Phương trình:

`C+O_2\overset{t^o}{\to}CO_2`

`S+O_2\overset{t^o}{\to}SO_2`

Ta nhận thấy: `n_{O_2}=x+y=0,3(mol)(2)`

Từ `(1),(2)` suy ra $\begin{cases} x=0,2(mol)\\ y=0,1(mol)\\\end{cases}$

`b)` `m_{C}=12.0,2=2,4g`

`m_{S}=32.0,1=3,2g`

`c)`Theo phương trình, ta nhận thấy:

`n_{C}=n_{CO_2}=0,2(mol)`

`n_S=n_{SO_2}=0,1(mol)`

`%m_C=\frac{2,4.100%}{5,6}\approx 42,86%`

`%m_S=\frac{3,2.100%}{5,6}\approx 57,14%`

`d)` `%n_{CO_2}=\frac{0,2.100%}{0,3}\approx 66,67%`

`%n_{SO_2}=\frac{0,1.100%}{0,3}\approx 33,33%`