a)
\(n_{CúO4}=\frac{100.16\%}{160}=0,1\left(mol\right)\)
\(n_{NaOH}=\frac{100.10\%}{40}=0,5\left(mol\right)\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,1______ 0,2____________0,1________0,1
\(m_{Cu\left(OH\right)2}=0,1.98=9,8\left(g\right)\)
b)
\(m_{dd.spu}=100+200-9,8=290,2\left(g\right)\)
\(C\%_{NaOH.du}=\frac{0,3.40}{290,2}.100\%=4,135\%\)
\(C\%_{Na2SO4}=\frac{0,1.42}{290,2}.100\%=4,89\%\)