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К раствору, содержащему 49 грамм серной кислоты, прилили избыток раствора нитрата бария. Найдите массу выпавшего осадка.
Ответы 1
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H2SO4 + Ba(NO3)2=BaSO4(ocaд.) + 2HNO3n(H2SO4)=n(BaSO4)n(H2SO4)=m/M=49/2+32+64=49/98=0.5 мольm(BaSO4)=n*M=0.5*(137+32+64)=116.5 грамм
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Автор:
danica - 6 лет назад
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NaNO3+H2SO4+Cu=NO2+Na2SO4+CuSO4+H20
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