60 г сульфида алюминия обработали избытком водного раствора хлороводородной кислоты. рассчитать объем(н у) газа, выделившегося в рез-те этой реакции.
valeriaurash20
3
22.05.2019 10:00
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1
Al2S3+6HCl=A2lCl3+3H2S
n(Al2S3)=60г/150г/моль=0,4моль
1моль-3моль
0,4 моль-Х
Х=1,2моль -n(H2S)
V(H2S)=1,2моль*22,4л/моль=26,88л