Bạn tham khảo:
$n_{NaOH}=\frac{25.4\%}{40}=0,025(mol)$
$n_{H_2SO_4}=0,2.0,052=0,0104(mol)$
$2NaOH+H_2SO_4 \to Na_2SO_4+H_2O$
$\frac{0,025}{2}=0,0125 > \frac{0,0104}{1}=0,0104$
$NaOH > H_2SO_4$
$H_2SO_4$ hết, $NaOH$ dư
$m_{dd}=25+51=76(g)$
$n_{Na_2SO_4}=n_{H_2SO_4}=0,0104(mol)$
$C\%_{Na_2SO_4}=\frac{0,0104.142}{76}.100\%=1,94\%$
$n_{NaOH}=0,0104.2=0,0208(mol)$
$C\%_{NaOH}=\frac{(0,025-0,0208).40}{76}.100\%=0,22\%$