1.a) \(A+HCl\rightarrow ACl+\dfrac{1}{2}H_2\)
Ta có : \(n_A=2n_{H_2}=2.\dfrac{2,8}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow M_A=\dfrac{9,75}{0,25}=39\)
Vậy A là Kali (K)
b)\(n_{HCl}=2n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,25.26,5=9,125\left(g\right)\)