PTHH: \(2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\)

a) \({n_{NaOH}} = 1,1mol;{n_{{H_2}S{O_4}}} = 0,5mol\)

 Nhận thấy: \(\dfrac{{{n_{NaOH}}}}{2} > \dfrac{{{n_{{H_2}S{O_4}}}}}{1}\)

Vậy \({H_2}S{O_4}\) hết → mol muối tạo thành tính theo \({H_2}S{O_4}\)

Theo PTHH: \({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,5mol \to {m_{N{a_2}S{O_4}}} = 0,5.142 = 71g\)

b) \({n_{NaOH}} = 0,4mol;{n_{{H_2}S{O_4}}} = 0,5mol\)

Khi đó \(\dfrac{{{n_{NaOH}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{1}\)

⇒\(NaOH\) hết ⇒ mol muối sinh ra tính theo \(NaOH\)

Theo PTHH: \({n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol \to {m_{N{a_2}S{O_4}}} = 0,2.142 = 28,4g\)

c) \({n_{NaOH}} = 0,8mol;{n_{{H_2}S{O_4}}} = 0,5mol\)

Có: \(\dfrac{{{n_{NaOH}}}}{2} < \dfrac{{{n_{{H_2}S{O_4}}}}}{1}\)

⇒\(NaOH\) hết ⇒ mol muối sinh ra tính theo \(NaOH\)

Theo PTHH: \({n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,4mol \to {m_{N{a_2}S{O_4}}} = 0,4.142 = 56,8g\)