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elena17051305y
@elena17051305y
July 2022
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13 ,пожалуйста, срочно!!!
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vasilinamolodan
H2SO4+2KOH=K2SO4+2H2OM(H2SO4)=98г/моль1л раствора содержит 9,8гр. кислоты0,1л раствора содержит 0,98гр. кислоты0,98гр. Хгр.H2SO4+2KOH=K2SO4+2H2O98гр. 112гр.Х=0,98*112/98=1,12гр.W(KOH)=1,12*100%/80=1,4%
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Alexei78
ДаноV=100млC(H2SO4)=0.1MmppaKOH=80gНайти W(KOH)-?Решение0.98. ХH2SO4+2KOH-->K2SO4+2H2O98. 2*56mвваH2SO4= C*M*V, M(H2SO4)=98г/мольm Вва= 0.1*98*0.1=0.98гМ(KOH)=56g/mol0.98/98=x/112. X=1.12gW(KOH)=mвва/mppa*100%=1.12/80*100=1.4%Ответ 1.4%
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