x=0.02 mol=0.04 g =0.448 l
y=0.18 mol= 6.57 g= 4.032 l
m(smesi)=6.61
V(smesi)= 4.48
ф%(H2)=0.448/4.48=0.1 =10% ф(HCl)=90%
w%(H2)=0.04/6.61=0.0006=0.6% w%(HCl)=99.4%
Объяснение:
x=0.02 mol=0.04 g =0.448 l
y=0.18 mol= 6.57 g= 4.032 l
m(smesi)=6.61
V(smesi)= 4.48
ф%(H2)=0.448/4.48=0.1 =10% ф(HCl)=90%
w%(H2)=0.04/6.61=0.0006=0.6% w%(HCl)=99.4%
Объяснение: