$a,n_{Zn=}$ $\dfrac{16,25}{65}=0,25(mol)$
$Zn+H_2SO_4→ZnSO_4+H_2$
$0,25→0,25$ $→0,25→0,25$
a,$m_{H_2SO_4}=0,25.98=24,5g$
$→C$%$_{H_2SO_4}=$$\dfrac{24,5}{100}.100$ %$=24,5$%
$V_{H_2}=0,25.22,4=5,6l$
$b,m_{ZnSO_4}=0,25.161=40,25g$
$m_{dd}$ sau pư$=16,25+100-0,25.2=115,75g$
$C$%$_{ZnSO_4}=$$\dfrac{40,25}{115,75}.100$%$=34,77$%