$n_{Zn} = \dfrac{13}{65} = 0,2(mol)$
$n_{H_2SO_4} = \dfrac{24,5}{98} = 0,25(mol)$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Ta thấy :
$n_{Zn} : 1 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư
$n_{ZnSO_4} = n_{H_2} = n_{H_2SO_4\ pư} = n_{Zn} = 0,2(mol)$
Suy ra :
$m_{ZnSO_4} = 0,2.161 = 32,2(gam)$
$m_{H_2} = 0,2.2 = 0,4(gam)$
$m_{H_2SO_4\ dư} = (0,25 - 0,2).98 = 4,9(gam)$
PTHH: Zn + H2SO4 -> a
Theo ĐLBTKL:
Theo pt: Zn + H2SO4 -> a
=> \(m_{Zn}+m_{H_2SO_4}=m_a\)
Thay 13 + 24,5 = 37,5 (g)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{24,5}{2+32+16\cdot4}=0,25\left(mol\right)\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ: 1 : 1 : 1 : 1
n(mol) 0,2 0,25
n(mol pu) 0,2---->0,2--------->0,2------->0,2
\(\dfrac{n_{Zn}}{1}< \dfrac{n_{H_2SO_4}}{1}\left(\dfrac{0,2}{1}< \dfrac{0,25}{1}\right)\)
=> Zn hết, H_2 SO_4` dư
=> tính theo Zn
\(n_{H_2SO_4}=0,25-0,2=0,05\left(mol\right)\\ m_{H_2SO_4}=n\cdot M=0,05\cdot\left(2+32+16\cdot4\right)=4,9\left(g\right)\\ m_{ZnSO_4}=n\cdot M=0,2\cdot\left(65+32+16\cdot4\right)=32,2\left(g\right)\\ m_{H_2}=n\cdot M=0,2\cdot2=0,4\left(g\right)\)
