n(H2S)=3.36/22.4=0.15mol
设有amolH2S与NaOH反应 有bmolH2S与Na2S反应
2NaOH+H2S=Na2S+2H2O
1 1
a x
x=amol
Na2S + H2S = 2NaHS
1 1 2
y b z
y=bmol z=2bmol
a+b=0.15
78(a-b)+2b*56=10.6
a=0.125mol
b=0.025mol
n(Na2S)=a-b=0.125-0.025=0.1mol
n(NaHS)=2b=0.025*2=0.05mol
m(Na2S)=0.1*78=7.8g
m(NaHS)=0.05*56=2.8g
固体中含有Na2S和NaHS
Na2S为7.8g; NaHS为2.8g