\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) => \(n_{H_2SO_4}=0,3mol\)
Mg + H2SO4 -------> MgSO4 + H2
Zn + H2SO4 -------> ZnSO4 + H2
Fe + H2SO4 -------> FeSO4 + H2
Ta có: mhh + \(m_{H_2SO_4}\)= mmuối + \(m_{H_2}\)
=> mmuối = 14,5 + 0,3.98 - 0,3.2
= 43,3 g
Mg + H2SO4 → MgSO4 + H2 (1)
Zn + H2SO4 → ZnSO4 + H2 (2)
Fe + H2SO4 → FeSO4 + H2 (3)
\(\Sigma n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,3\times2=0,6\left(g\right)\)
Theo PT1,2,3: \(\Sigma n_{H_2SO_4}=\Sigma n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,3\times98=29,4\left(g\right)\)
Áp dụng ĐL BTKL: \(m_{hh}+m_{HCl}=m_{m'}+m_{H_2}\)
\(\Rightarrow m_{m'}=m_{hh}+m_{HCl}-m_{H_2}=14,5+29,4-0,6=43,3\left(g\right)\)
nH2 = \(\dfrac{6,72}{22,4}\)= 0,3 mol
Mg + H2SO4 -> MgSO4 + H2
Zn + H2SO4 -> ZnSO4 + H2
Fe + H2SO4 -> FeSO4 + H2
ta có :
mM = mKL + mSO4 \(\Leftrightarrow\)mM = mKL + nH2 . 96
\(\Leftrightarrow\)mM = 14,5 + 0,3.96 = 43,3 g
vậy khối lượng muối khan thu đc sau phản ứng là 43,3g