Câu 10:
a, PT: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_{2\downarrow}\)
b, Ta có: \(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
Theo PT: \(n_{Na_2SO_4}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
Ta có: m dd sau pư = 200 + 100 - 0,25.98 = 275,5 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,25.142}{275,5}.100\%\approx12,89\%\)
c, PT: \(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)
Ta có: \(n_{HCl}=0,2.3=0,6\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,6}{2}\), ta được HCl dư.
Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,25.135=33,75\left(g\right)\)
