a,
$Mg+2HCl\to MgCl_2+H_2$
b,
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$\Rightarrow n_{Mg}=0,1(mol)$
$m_{Mg}=0,1.24=2,4g$
$\Rightarrow m_{Cu}=8,8-2,4=6,4g$
Sau phản ứng, dd còn $6,4g$ chất rắn.
c,
$n_{HCl}=2n_{H_2}=0,2(mol)$
$\Rightarrow m_{dd HCl}=0,2.36,5:19,87\%=36,74g$
$\Rightarrow V_{dd HCl}=\dfrac{36,74}{1,047}=35(ml)$