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rostikpupkin
@rostikpupkin
July 2022
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Какой будет объем водорода при , взаимодействии 54 грам аллюминия с HCl
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DanToWiN
2Al+6HCl=2AlCl3+3H2
n(Al)=m/M=54/27=2 моль
V(H2)=n*V молярный=3*22.4= 67.2л
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Answers & Comments
n(Al)=m/M=54/27=2 моль
V(H2)=n*V молярный=3*22.4= 67.2л