\(4Na+O_2\rightarrow2Na_2O\left(1\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\) (2)
\(2Cu+O_2\rightarrow2CuO\) (3)
\(Na_2O+2HCl\rightarrow2NaCl+H_2O\) (4)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (5)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\) (6)
Áp dụng ĐLBTKL ở pthh 1,2,3 ta có:
\(m_{O\left(h^2B\right)}=12,2-m\left(g\right)\)
\(\Rightarrow n_{O\left(h^2B\right)}=0,7625-0,0625m\left(mol\right)\)
Theo PTHH 4,5,6: \(n_{HCl}=2n_{O\left(trongB\right)}=1,525-0,125m\left(mol\right)\)
\(\Rightarrow m_{HCl}=55,6625-4,5625m\left(g\right)\)
\(n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,7625-0,0625m\left(mol\right)\)
Theo ĐLBTKL ở pthh 4,5,6:
\(m_B+m_{HCl}=m_{muối}+m_{H_2O}\)
\(\Rightarrow12,2+55,6625-4,5625m=25,95+13,725-1,125m\)
\(\Rightarrow m=8,2\)
Vậy mA = 8,2g.