Đáp án:

`↓`

Giải thích các bước giải:

PTHH: `CuSO_4 + 2NaOH -> Cu(OH)_2↓ + Na_2 SO_4`

`mCuSO_4 = (300 . 12)/100 = 36 (g)`

`-> nCuSO_4 = 36/160 = 0,225 (mol)`

`a.`

Theo PT: `nNaOH = nCuSO_4 . 2 = 0,225 . 2 = 0,45 (mol)`

`mNaOH = 0,45 . 40 = 18 (g)`

`-> C%_{ddNaOH} = (18 . 100)/200 = 9 (%)`

`b.`

Theo PT: `nCu(OH)_2 = nCuSO_4 = 0,225 (mol)`

`-> mCu(OH)_2 = 0,225 . 98 = 22,05 (g)`

`c.`

Theo PT: `nNa_2 SO_4 = nCuSO_4 = 0,225 (mol)`

`-> mNa_2 SO_4 = 0,225 . 142 = 31,95 (g)`

`mddsaupu = mddCuSO_4 + mddNaOH - mCu(OH)_2`

`-> mddsaupu = 300 + 200 - 22,05 = 477,95 (g)`

`-> C%_{ddNa_2 SO_4} = (31,95 . 100)/(477,95) = 6,68 (%)`

Đáp án:

$a,\ C\%_{dd\ NaOH}=9\%\\ b,\ m_{Cu(OH)_2}=22,05(g)\\ c,\ C\%_{dd\ Na_2SO_4}=6,68\%$

Giải thích các bước giải:

$a,\ n_{CuSO_4}=\dfrac{300.12\%}{160}=0,225(mol)\\ CuSO_4+2NaOH\to Cu(OH)_2\downarrow+Na_2OS_4\\ n_{NaOH}=2.n_{CuSO_4}=0,45(mol)\\ m_{NaOH}=0,45.40=18(g)\\ C\%_{dd\ NaOH}=\dfrac{18}{200}.100\%=9\%\\ b,\ n_{Cu(OH)_2}=n_{CuSO_4}=0,225(mol)\\ m_{Cu(OH)_2}=0,225.98=22,05(g)\\ c,\ n_{Na_2SO_4}=n_{CuSO_4}=0,225(mol)\\ m_{Na_2SO_4}=0,225.142=31,95(g)\\ m_{dd\ Na_2SO_4}=300+200-22,05=477,95(g)\\ C\%_{dd\ Na_2SO_4}=\dfrac{31,95}{477,95}.100\%=6,68\%$