Đáp án:

 

Giải thích các bước giải:

$g\bigg)$

$n_{CO_2}=\frac{7,056}{22,4}=0,315(mol)$

Đặt CTPT ankan là $C_nH_{2n+2}$

Bảo toàn C: $n_{ankan}.n=n_{CO_2}=0,315$

$\to \frac{4,5n}{14n+2}=0,315$

$\to n=7$

$\to CTPT:C_7H_{16}$

$CH_3-CH_2-CH_2-CH_2-CH_2-CH_2-CH_3$

$CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3$

$CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$

$CH_3-CH(CH_3)-CH(CH_3)-CH_2-CH_3$

$CH_3-CH(CH_3)-CH_2-CH(CH_3)-CH_3$

$CH_3-C(CH_3)_2-CH_2-CH_2-CH_3$

$CH_3-C(CH_3)_2-CH(CH_3)-CH_3$

$CH_3-CH_2-C(CH_3)_2-CH_2-CH_3$

$CH_3-CH_2-CH(CH_2)(CH_3)-CH_2-CH_3$

$h\bigg)$

$n_{H_2O}=\frac{22,5}{18}=1,25(mol)$

Đặt CTPT ankan là $C_nH_{2n+2}$

Bảo toàn H: $n_{ankan}.(n+1)=n_{H_2O}$

$\to \frac{14,5(n+1)}{14n+2}=1,25$

$\to n=4$

$\to CTPT:C_4H_{10}$

$CH_3-CH_2-CH_2-CH_3$

$CH_3-CH(CH_3)-CH_3$

$i\bigg)$

$n_{O_2}=\frac{3,64}{22,4}=0,1625(mol)$

Đặt CTPT ankan là $C_nH_{2n+2}$

$C_nH_{2n+2}+\frac{3n+1}{2}O_2\xrightarrow{t^o}nCO_2+(n+1)H_2O$

Theo PT: $n_{ankan}=\frac{2n_{O_2}}{3n+1}=\frac{0,325}{3n+1}$

$\to \frac{1,45}{14n+2}=\frac{0,325}{3n+1}$

$\to n=4$

$\to CTPT:C_4H_{10}$

$CH_3-CH_2-CH_2-CH_3$

$CH_3-CH(CH_3)-CH_3$