PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\) (2)
\(n_{HCl\left(1\right)+\left(2\right)}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
\(n_{H_2}=\dfrac{7,437}{24,79}0,3\left(mol\right)\)
Theo PTHH (1): \(n_{Al}=\dfrac{2\cdot0,3}{3}=0,2\left(mol\right)\); \(n_{HCl\left(1\right)}=\dfrac{6\cdot0,3}{3}=0,6\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2\cdot27=5,4\left(g\right)\)
\(\Rightarrow n_{HCl\left(2\right)}=1-0,6=0,4\left(mol\right)\)
Theo PTHH (2): \(n_{CuO}=\dfrac{1\cdot0,4}{2}=0,2\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)
\(\Rightarrow m_A=m_{CuO}+m_{Al}=16+5,4=21,4\left(g\right)\)
% khối lượng của mỗi chất trong hỗn hợp A là:
\(\%m_{Al}=\dfrac{5,4\cdot100}{21,4}\approx25,2\%\)
\(\Rightarrow\%m_{CuO}=100\%-25,2\%=74,8\%\)
