Đáp án:

$$(NH_4)_2SO_4+2NaOH\to Na_2SO_4+2NH_3+2H_2O$$ 

$$NH_3+HCl\to NH_4Cl$$ 

$$HCl+NaOH\to NaCl+H_2O$$ 

$$n_{NaOH}=11,10.0,1210=1,343$$ mmol 

$$\Rightarrow n_{HCl\text{dư}}=1,343$$ mmol 

$$n_{HCl}=50,0.0,1000=5,00$$ mmol 

$$\Rightarrow n_{HCl\text{pứ}}=n_{NH_3}=5,00-1,343=3,66$$ mmol 

$$\%NH_3=\dfrac{3,66.17,03.10^{-3}.100\%}{0,4750}=13,1\%$$ 

$$n_{(NH_4)_2SO_4}=\dfrac{3,66}{2}=1,83$$ mmol 

$$\%(NH_4)_2SO_4=\dfrac{1,83.132,1.10^{-3}.100\%}{0,4750}=50,9\%$$