m(прим)=120×0,11=13,2г
m(KOH)=120-13,2=106,8г
n(KOH)=106,8:39≈1,9моль
1,9. x. x
2KOH+H2-->2K+2H2O
x(H2)=1,9×22,4=42,56л
x(H2O)=(1,9×2)×18=68,4г
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m(прим)=120×0,11=13,2г
m(KOH)=120-13,2=106,8г
n(KOH)=106,8:39≈1,9моль
1,9. x. x
2KOH+H2-->2K+2H2O
x(H2)=1,9×22,4=42,56л
x(H2O)=(1,9×2)×18=68,4г