a/ $2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
b/ $n_{NaOH}=\dfrac{16}{40}=0,4(mol)$
Theo PTHH: $n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}$
$→n_{H_2SO_4}=0,2(mol)\\→m_{H_2SO_4}=0,2.98=19,6g\\m_{dd\,H_2SO_4}=\dfrac{19,6}{9,8\%}=200g$
Vậy $m_{dd\,H_2SO_4}=200g$
c/ $m_{dd\,\text{sau phản ứng}}=16+200=216g$
Theo PTHH: $n_{Na_2SO_4}=n_{H_2SO_4}$
$→n_{Na_2SO_4}=0,2(mol)\\→m_{Na_2SO_4}=0,2.142=28,4g$
$C\%_{Na_2SO_4}=\dfrac{28,4}{200}=14,2\%$
Vậy $C\%_{Na_2SO_4}=14,2\%$