`#DyHugg`

Bài `1` :

`SO_2+Ca(OH)_2->CaSO_3↓+H_2O`

`a)`

`->n_{SO_2}=(7,84)/(22,4)=0,35\ (mol)`

`->n_{Ca(OH)_2=0,35\ (mol)`

`->C_{M(Ca(OH)_2)}=(0,35)/(0,25)=1,4\ (M)`

`b)`

`n_{CaSO_3}=0,35\ (mol)`

`->m_{CaSO_3}=0,35.120=42\ (g)`

Bài `2` :

`a)`

PTHH: `CuO+2HCl->CuCl_2+H_2O`

`n_{CuO}=10/80=0,125\ (mol)`

`b)`

`n_{HCl}=2n_{CuO}=0,125.2=0,25\ (mol)`

`->m_{HCl}=0,25.36,5=9,125\ (g)`

`c)`

`n_{CuCl_2}=0,125\ (mol)`

`->C_{M}=(0,125)/(0,5)=0,25 (M)`

Đáp án + giải thích các bước giải:

Bài `1:`

`a.`

`n_{SO_2}=\frac{7,84}{22,4}=0,35(mol)`

Phương trình hóa học :

`SO_2+Ca(OH)_2 \to CaSO_3↓+H_2O`

$\xrightarrow[]{\text{Theo PTHH}}n_{Ca(OH)_2}=n_{SO_2}=0,35(mol)$

`V_{Ca(OH)_2}=250(ml)=0,25(l)`

` \to CM_{Ca(OH)_2}=\frac{0,35}{0,25}=1,4(M)`

`b.`

Muối thu dược `:CaSO_3`

$\xrightarrow[]{\text{Theo PTHH}}n_{CaSO_3}=n_{SO_2}=0,35(mol)$

` \to m_{\text{muối}}=m_{CaSO_3}=0,35.120=42(g)`

Bài `2:`

`a.`

Phương trình hóa học :

`CuO+2HCl \to CuCl_2+H_2O`

`b.`

`n_{CuO}=\frac{10}{80}=0,125(mol)`

$\xrightarrow[]{\text{Theo PTHH}}n_{HCl}=2.n_{CuO}=2.0,125=0,25(mol)$

`m_{HCl}=0,25.36,5=9,125(g)`

`c.`

$\xrightarrow[]{\text{Theo PTHH}}n_{CuCl_2}=n_{CuO}=0,125(mol)$

`V_{CuCl_2}=v_{HCl}=500(ml)=0,5(l)`

` \to CM_{CuCl_2}=\frac{0,125}{0,5}=0,25(M)`