Предмет: Химия, автор: Sofaka1234

определить объем водорода, который можно получить действуя на 108 г алюминия соляной кислотой ,если объемная доля выхода продукта составляет 85 %


Плиииз!!!

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Автор ответа: affone4ka
0
2Al+6HCl = 2AlCl3+3H2
 nAl = m/M = 108/27 = 4 моль 
 nH2 = 3/2*nAl = 6 моль (теоретически)
nH2 = 0.85*6 = 5.1 моль (по факту)
VH2 = Vm*n = 22.4*5.1m = 114.24л                                  
Автор ответа: Sofaka1234
0
спасибо!!!
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