a)\(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
\(C_{M_{NaOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
b) \(n_{\left(50ml\right)NaOH}=2.0,05=0,1\left(mol\right)\)
\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT:.......2 mol............1mol..........1mol...........2mol...
Theo ĐB:......0,1mol...........\(n_{H_2SO_4}\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{4,9.100}{19,6}=25\left(g\right)\)