Дано
m(ppaH2SO4)=40 g
W(H2SO4)=10%
η(CuSO4)=82%
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m(Cu(OH)2)-?
m(CuSO4)-?
m(H2SO4)=40*10%/100%=4 g
X 4 Y
Cu(OH)2+H2SO4-->CuSO4+2H2O
97.5 98 159.5
M(Cu(OH)2)=97.5 g/mol
M(H2SO4)=98 g/mol
M(CuSO4)=159.5 g/mol
X=97.5*4 / 98 ≈ 4 g
Y=4*159.5 / 98
Y=6.51 g
m(CuSO4)=6/51*82% / 100% = 5.34 g
ответ 4 г- Cu(OH)2 , 5.34 г- CuSO4