$a)\\C_nH_{2n+2} + \frac{3n+1}{2}O_2 \xrightarrow{t^o} nCO_2 + (n+1)H_2O\\→\frac{29}{14n+2}=\frac{1}{n}×\frac{88}{44}\\→29n=28n+4 →n=4\\→\text{CTPT là: }C_4H_{10}\\\text{Đồng phân: Trong hình}\\b)\\C_4H_{10} + \frac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O\\n_{O_2}=\frac{13}{2}n_{C_4H_{10}}=\frac{13}{2}×\frac{5,8}{58}=0,65\,(mol)\\→V_{O_2}=0,65×22,4=14,56\,(l)\\→V_{KK}=5V_{O_2}=72,8\,(l)$