al2(so4)3+3bacl2=2alcl3+3baso4m(чистая al2(so4)3)=68,4*0,08=5,472гn(al2(so4)3)=5,472г / 342 г/моль=0,016 моль n(baso4)=3*n(al2(so4)3)=3*0,016=0,048моль m(baso4)=0,048моль*233 г/моль=11,184 гответ: m(baso4)=11,184г
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Объяснение:
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Поделитесь своими знаниями, ответьте на вопрос:
Краствору сульфата алюминия массой 68, 4г и массовой долей 8% прилили избыток раствора хлорида бария. вычислите массу образовавшегося осадка
m (al2(so4)3)=m(p-pa)*ω(al2(so4)3)/100%=68,4 г * 0,08=5,472г
n (al2(so4)3)=m/m=5,472 г / 342 г/моль=0,016 моль
0,016 моль х
3 bacl2+al2(so4)3=2 alcl3+3 baso4↓
1 моль 3 моль
х=0,048моль
m (baso4)=n*m=0,048 моль * 233 г/моль=11,184г