к серной кислоте прилили раствор , содержащий 10.4 хлорида бария. Определите массу осадка
H
n=m/M
n(BaCl_{2})=10.4/207=0.05
n(BaCl_{2})=n(BaSO_{4})=0.05
m=n*M
m(BaSO_{4})=0.05*233=11.65
Автор:
flakehperH2SO4 + BaCl2->BaSO4 + 2HCl
10.4г (BaCl2) - Xг(BaSO4)
208г/м - 233г/м
X=11,65г
если 10,4 л BaCl2, то вместо 208 подставь 22,4л/м
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