a)\(Zn+Cl2-->ZnCl2\)
b)\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(n_{Cl2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
\(\Rightarrow Zndư\)
Chất rắn A là Zn dư
\(n_{Zn}=nCl2=0,06\left(mol\right)\)
\(n_{Zn}dư=0,1-0,06=0,04\left(mol\right)\)
\(m_A=m_{Zn}dư=0,04.65=2,6\left(g\right)\)