Đáp án `+` Giải thích các bước giải:

$*$ $H_2$ : 

 `-` $M_{H_2}=1×2=2(g/mol)$

 `-` $n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\dfrac{4}{2}=2(mol)$ 

 `-` $V_{H_2}=n_{H_2}×22,4=2×22,4=44,8(l)$ 

$*$ $N_2$ : 

 `-` $M_{N_2}=14×2=28(g/mol)$

 `-` $n_{N_2}=\dfrac{m_{N_2}}{M_{N_2}}=\dfrac{2,8}{28}=0,1(mol)$ 

 `-` $V_{N_2}=n_{N_2}×22,4=0,1×2,24=2,24(l)$ 

$*$ $O_2$ : 

 `-` $M_{O_2}=16×2=32(g/mol)$

 `-` $n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{6,4}{32}=0,2(mol)$ 

 `-` $V_{O_2}=n_{O_2}×22,4=0,2×22,4=4,48(l)$ 

$*$ $CO_2$ : 

 `-` $M_{CO_2}=12+16×2=44(g/mol)$

 `-` $n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{22}{44}=0,5(mol)$ 

 `-` $V_{CO_2}=n_{CO_2}×22,4=0,5×22,4=11,2(l)$ 

`⇒` Chọn $C$

$#Kiều$