Bài 4 :
$n_{Al_2O_3} = \dfrac{20,4}{102} = 0,2(mol)$
$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
Theo PTHH :
$n_{H_2SO_4} = 3n_{Al_2O_3} = 0,6(mol)$
$C\%_{H_2SO_4} = \dfrac{0,6.98}{112,504} .100\% = 52,26\%$
$m_{dd\ sau\ pư} = 20,4 + 112,504 = 132,904(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,2.342}{132,904}.100\% = 51,5\%$
Bài 5 :
a) $BaO + H_2SO_4 \to BaSO_4 + H_2O$
$BaO + H_2O \to Ba(OH)_2$
Theo PTHH : $n_{BaSO_4} = n_{H_2SO_4} = 0,02(mol)$
$m_{BaSO_4} = 0,02.233 = 4,66(gam)$
b) $C\%_{H_2SO_4} = \dfrac{0,02.98}{101,5}.100\% = 1,93\%$
Đúng 3
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