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1 . Fe+2HCl=FeCl2+H2
1) n(Fe)=m( Fe)/M( Fe)=25/56=0.4464
2)n( Fe)=n(H2)=0.4464
3) V(H2)=n(H2)*V=22.4*0.4464=10
2. CuO+H2SO4=CuSO4+H2O
1) n(CuO)=m(CuO)/M(CuO)=20/80=0.25
2) n(CuO)=n(CuSO4)=0.25
3) m(CuSO4)=M(CuSO4)*n(CuSO4)=160*0.25=40
3. CO2+Ca(OH)2=CaCO3+H2O
CaCO3+H2O+CO2=Ca(HCO3)2
1) n(Ca(HCO3)2)=m(Ca(HCO3)2)/M(Ca(HCO3)2=8.1/162=0.05
2) если для двух реакции нам нужно одинаковые моли CO2 то мы должны умножить на 2
3) n(CO2)=2n(Ca(HCO3)2)=0.05*2=0.1
4) V(CO2)=n(CO2)*V=0.1*22.4=2.24
4. FeS+2HCl=FeCl2+H2S
1)n(Fes)=n(H2S)=0.3
2) V(H2S)=V*n(H2S)=0.3*22.4=6.72
1. Вообщем для того, чтобы вычислить массовую долю вещества. Делишь массу растворенного вещества на массу раствора.
Масса растворенного вещества - это масса аммиака, который растворили.
n(NH3)=V/Vm=112/22,4=5 моль
m(NH3)=nM=5*17=85 г
Теперь находим массу всего раствора.
300+85=385
А теперь все просто делим.
W(NH3)=85/385=0,22*100%=22%
ответ: 22%
2. Задачка аналогична.
2NH3+H2SO4=(NH4)2SO4
n(H2SO4)=n((NH4)2SO4)=6,6/132=0,05 моль
m(H2SO4)=nM=0,05*98=4,9г
W(H2SO4)=4,9/20=0,245*100%=24,5%
ответ: 24,5%