Bài 1:
a,
$2A+2aHCl\to 2ACl_a +aH_2$
$2B+2bHCl\to 2BCl_b + bH_2$
b,
$n_{H_2}=0,4 mol$
Theo PTHH, $n_{HCl}=2n_{H_2}= 0,8 mol$
Bảo toàn khối lượng:
$a= m_{\text{muối}}+m_{H_2}-m_{HCl}$
$= 67+0,4.2-0,8.36,5=38,6g$
Bài 2:
$m_{CuSO_4}=500.5\%= 25g$
$\Rightarrow n_{CuSO_4.5H_2O}= n_{CuSO_4}=\frac{25}{160}=0,15625 mol$
$m_{CuSO_4.5H_2O}=0,15625.250=39,0625g$
$m_{H_2O}=500-39,0625=460,9375g$