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@271076
October 2021
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Вычислите объем газа(н.у), который выделится при действии избытка меди на 100 г 98-% раствора серной кислоты
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kuhta1199
Verified answer
Cu+2H2SO4=CuSO4+SO2+2H2O
m(H2SO4)=0,98×100=98г
n(H2SO4)=98/98=1моль
n(SO2)=1/2n(H2SO4)=1×0,5=0,5моль
V(SO2)=22,4×0,5=11,2л
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Answers & Comments
Verified answer
Cu+2H2SO4=CuSO4+SO2+2H2Om(H2SO4)=0,98×100=98г
n(H2SO4)=98/98=1моль
n(SO2)=1/2n(H2SO4)=1×0,5=0,5моль
V(SO2)=22,4×0,5=11,2л