PTHH:
2X + 2mHCl ---> 2XClm (A) + mH2
2Y + 2nHCl ---> 2YCln (B) + nH2
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Bảo toàn H: \(n_{HCl}=2n_{H_2}=2.0,6=1,2\left(mol\right)\)
\(\rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
Bảo toàn khối lượng:
\(m_{muối\left(A,B\right)}=27,2+43,8-0,6.2=69,8\left(g\right)\)