đặt nCu = a (mol) ; nMg = b (mol Cu + 1/2 O2 -> CuO
(mol) a -> a
Mg + 1/2 O2 -> MgO
(mol) b -> b
m (hỗn hợp) = mCu + mMg = 64a + 24 b = 7.6 (1)
%MgO = 20% => %CuO = 80%
mMgO/ %MgO = mCuO / %CuO
=> 40b / 20 = 80a / 80
=> 2b = a (2)
(1) (2) => a= 0.1 (mol) ; b = 0.05 (mol)
mCu = 6.4g ; mMg = 1.2g
CuO + 2HCl -> CuCl2 + H2O
0.1 -> 0.2
MgO + 2HCl -> MgCl2 + H2O
0.05 -> 0.1
=> nHCl = 0.3 (mol)
VddHCl = n / CM = 0.3 / 0.5 = 0.6 M