a)
$n_{Zn} = \dfrac{13}{65} = 0,2(mol) ; n_{H_2 SO_4} = 0,5.2 = 1(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy :
$n_{Zn} < n_{H_2SO_4}$ nên $H_2SO_4$ dư
$n_{ZnSO_4} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
$m_{ZnSO_4} = 0,2.161=32,2(gam)$
$m_{H_2SO_4\ pư} = 0,2.98 = 19,6(gam)$
b)
$n_{H_2SO_4\ dư} = 1 - 0,2 = 0,8(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,8}{0,5} = 1,6M$
$C_{M_{FeSO_4}} = \dfrac{0,2}{0,5} = 0,4M$
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2SO_4}=0,5.2=1\left(mol\right)\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.Vì:\dfrac{0,2}{1}< \dfrac{1}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\\ m_{H_2SO_4\left(p.ứ\right)}=0,2.98=19,6\left(g\right)\\ m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ b.V_{ddsau}=V_{ddH_2SO_4}=0,5\left(l\right)\\ C_{MddZnSO_4}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\ C_{MddH_2SO_4\left(dư\right)}=\dfrac{1-0,2}{0,5}=1,6\left(M\right)\)
