Đáp án:

Giải thích các bước giải:

`Bài 1:`

`a) PTHH:2H_2+O_2->2H_20`

`b)` Có `n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{1,12}{22,4}=0,05(mol)`

Theo `PTHH:n_{O_2}=2.n_{H_2}=2.0,05=0,1(mol)`

`⇒V_{O_2}=0,1.22,4=2,24(l)`

`c)` Theo `PTHH:`

`n_{H_2}=n_{H_2O}=0,05(mol)`

`⇒m_{H_2O}=n_{H_2O}.M_{H_2O}=0,05.(2.1+16)=0,05.18=0,9(g)`

Bài 2:

`a)` Có `PTHH:H_2+CuO->Cu+H_2O`

`b)` Có `n_{CuO}=\frac{m_{CuO}}{M_{CuO}}=\frac{12}{64+16}=\frac{12}{80}=0,15(mol)`

Theo `PTHH:n_{H_2}=n_{CuO}=0,15(mol)`

`⇒V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36(l)`

`c)` Theo `PTHH:`

`n_{CuO}=n_{Cu}=0,15(mol)`

`⇒m_{Cu}=n_{Cu}.M_{Cu}=0,15.64=9,6(g)`

Đáp án + Giải thích các bước giải:

BT1:

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

$b)n_{H_2}=\frac{1,12}{22,4}=0,05(mol)$

$n_{O_2}=\frac{1}{2}n_{H_2}=0,025(mol)$

$V_{O_2(đktc)}=0,025.22,4=0,56(l)$

$c)n_{H_2O}=n_{H_2}=0,05(mol)$

$m_{H_2O}=0,05.18=0,9(g)$

BT2:

$a)CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$b)n_{CuO}=\frac{12}{80}=0,15(mol)$

$n_{H_2}=n_{CuO}=0,15(mol)$

$V_{H_2(đktc)}=0,15.22,4=3,36(l)$

$c)n_{Cu}=n_{CuO}=0,15(mol)$

$m_{Cu}=0,15.64=9,6(g)$