Question

Answer 1

a) Gọi $n_{CuO} = a(mol) ; n_{ZnO} = b(mol) \Rightarrow 80a + 81b = 12,1(1)$
$CuO +2 HCl \to CuCl_2 + H_2O$
$ZnO + 2HCl \to ZnCl_2 + H_2O$

Theo PTHH : $n_{HCl} = 2a + 2b = 0,1.3 = 0,3(2)$

Từ (1)(2) suy ra : a = 0,05 ; b = 0,1

$\%m_{CuO} = \dfrac{0,05.80}{12,1}.100\% = 33,06\%$

$\%m_{ZnO} = 100\% - 33,06\% = 66,94\%$

b)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$ZnO + H_2SO_4 \to ZnSO_4 + H_2O$
$n_{H_2SO_4} = 0,05 + 0,1 = 0,15(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,15.98}{20\%} = 73,5(gam)$