Гидроксид натрия массой 49 г реагирует с 49 г серной кислоты. Масса образовавшегося сульфата натрия
Дано:
m(NaOH) = 49г
m(H2SO4)=49г
Найти: m(Na2SO4)=?
Решение:
n(NaOH) = m(NaOH)/M(NaOH)=48/40=1.2моль
n(H2SO4)=m(H2SO4)/M(H2SO4)=49/98=0.5моль
(Мне кажется что с условием что то напутано)
n(H2SO4)=n(Na2SO4)=0.5 моль
m(Na2SO4)=n(Na2SO4)*M(Na2SO4) =0.5*142=71 г
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