Определите массу железа (Fe) и объем кислорода (02) (при н.у.), которые были затрачены на образование 150 г оксида железа (|||), в реакциии Fe+O2O3 (перед решениме реакция уровнять)
DarinaDoka666
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14.11.2021 14:34
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Відповідь:
4Fe+3O2=2Fe2O3
M(Fe2O3)=160г/моль
M(Fe)=56г/моль
n(Fe2O3)=150/160=0,94моль
n(Fe):n(O2):n(Fe2O3)=4:3:2
n(Fe)=0,94*2=1,84моль
m(Fe)=1,84*56=52,5г
n(O2)=0,94*3/4=0,705моль
V(O2)=22,4*0,705=15,8л
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