a,
$n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1(mol)$
$V_{O_2}=11,2.20\%=2,24l$
$\to n_{O_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$C_2H_5OH+3O_2\xrightarrow{{t^o}} 2CO_2+3H_2O$
$\to O_2$ hết, $C_2H_5OH$ dư
Vậy hỗn hợp khí, hơi sau phản ứng gồm $C_2H_5OH, CO_2, H_2O, N_2$
b,
$n_{C_2H_5OH\text{pứ}}=\dfrac{0,1}{3}=\dfrac{1}{30}(mol)$
$\to n_{C_2H_5OH\text{dư}}=0,1-\dfrac{1}{30}=\dfrac{1}{15}(mol)$
$V_{N_2}=11,2.80\%=8,96l$
$n_{CO_2}=\dfrac{1}{30}.2=\dfrac{1}{15}(mol)$
$n_{H_2O}=\dfrac{1}{30}.3=0,1(mol)$
$\to V_2=22,4\Big( \dfrac{1}{15}+\dfrac{1}{15}+0,1\Big)+8,96=14,19(l)$