Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
a. \(PTHH:CuSO_4+2NaOH--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
b. Theo PT: \(n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)
c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)
\(\Rightarrow C_{\%_{NaOH}}=\dfrac{8}{200}.100\%=4\%\)