cho dd chứa 16g CuSO4 tác dụng vừa đủ với 200g dd NaOH. Viết PTHH . tính khối lượng chất kết tủa tạo thành . Tính nồng độ % của dd NaOH giúp e vs ạ

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Mời bạn tham khảo !!!

Đáp án:

PTHH : `CuSO_4 + 2NaOH -> Cu(OH)_2 + Na_2SO_4`

Chất kết tủa là : `Cu(OH)_2`

`n_{CuSO_4} = (16)/(160) = 0,1` `mol`

`n_{Cu(OH)_2} = n_{CuSO_4} = 0,1` `mol`

`m_{Cu(OH)_2} = 0,1 . 98 = 9,8` `gam`

`n_{NaOH} = 2 . n_{CuSO_4} = 0,2` `mol`

`m_{NaOH} = 0,2 . 40 = 8` `gam`

`C%_{NaOH} = (m_(ct))/(m_(dd)) . 100% = 8/(200) . 100% = 4%`

Bạn tham khảo!

 $\text{Ta có PTHH sau:}$

$\text{CuSO4+2NaOH$\rightarrow$Na2SO4+Cu(OH)2↓}$

$\text{Số mol của CuSO4 là:}$

$\text{n=$\frac{m}{M}$=$\frac{16}{160}$=0,1 mol}$

$\text{Có nCu(OH)2=nCuSO4=0,1 mol}$

$\rightarrow$ $\text{mCu(OH)2=n.M=0,1.98=9,8g}$

$\text{Có nNaOH=2nCuSO4=0,1.2=0,2 mol}$

$\rightarrow$ $\text{mNaOH=0,2.40=8g}$

$\text{C%NaOH=$\frac{mct}{mdd}$.100=$\frac{8}{200}$.100=4%}$

#TRANPHAMGIABAO

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