PTHH: AO + 2HCl → ACl2 + H2O
a) \(n_{AO}=\dfrac{24,3}{A+16}\left(mol\right)\)
\(n_{ACl_2}=\dfrac{40,8}{A+71}\left(mol\right)\)
Mà theo PT: \(n_{AO}=n_{ACl_2}\)
\(\Rightarrow\dfrac{24,3}{A+16}=\dfrac{40,8}{A+71}\)
\(\Rightarrow24,3A+1725,3=40,8A+652,8\)
\(\Leftrightarrow1072,5=16,5A\)
\(\Leftrightarrow A=65\)
Vậy A là nguyên tố kẽm Zn
b) PTHH: ZnO + 2HCl → ZnCl2 + H2O
\(n_{ZnO}=\dfrac{24,3}{81}=0,3\left(mol\right)\)
Theo PT: \(n_{HCl}pư=2n_{ZnO}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}pư=0,6\times36,5=21,9\left(g\right)\)