Đáp án + Giải thích các bước giải:

Câu 1:

$n_{H_2O}=\frac{3,6}{18}=0,2(mol)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{H_2}=n_{H_2O}=0,2(mol)$

$\to V_{H_2}=0,2.22,4=4,48(l)$

Câu 2:

$a\big)2Zn+O_2\xrightarrow{t^o}2ZnO$

$b\big)$

$n_{Zn}=\frac{19,5}{65}=0,3(mol)$

Theo PT: $n_{ZnO}=n_{Zn}=0,3(mol)$

$\to m_{ZnO}=0,3.81=24,3(g)$

$c\big)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{Zn}=0,15(mol)$

$\to V_{O_2}=0,15.22,4=3,36(l)$

$\to V_{không\,khí}=\frac{3,36}{20\%}=16,8(l)$

Giải thích các bước giải:

Câu `1.`
$\rm PTHH:2H_2+O_2\to2H_2O$
`n_(H_2O)=(m_(H_2O))/(M_(H_2O))=(3,6)/(1.2+16)=0,2(mol)`
Theo PTHH:`n_(H_2)=n_(H_2O)=0,2(mol)`
`->V_(H_2(\text{đktc}))=n_(H_2) . 22,4=0,2 . 22,4=4,48 (l)`
Câu `2.`
`a)`
$\rm PTHH:2Zn+O_2\to 2ZnO $
`b)`
`n_(Zn)=(m_(Zn))/(M_(Zn))=(19,5)/(65)=0,3(mol)`
Theo PTHH:`n_(ZnO)=n_(Zn)=0,3(mol)`
`->m_(ZnO)=n_(ZnO).M_(ZnO)=0,3 . (65+16)=24,3(g)`
`c)`
Theo PTHH:`n_(O_2)=1/2 . n_(Zn)=1/2 . 0,3=0,15(mol)`
`->V_(O_2(text{đktc}))=n_(O_2).22,4=0,15 . 22,4=3,36(l)`
`->V_(text{không khí})=3,36 :20 . 100=16,8(l)`