Question

If 0.436 moles of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate would be produced in the following reaction?
Pb(SO4)2 +2Zn ——> 2ZnSo4 +Pb

Answer 1

The balanced equation for reaction of zinc reacts with lead (IV) sulfate can be written as follows

$Pb(SO_{4})_{2}+2Zn\rightarrow2ZnSO_{4}+Pb$

$\frac{\nu(Zn)}{N(Zn)}=\frac{m(ZnSO_{4})}{N(ZnSO_{4})\cdot M(ZnSO_{4})}$

$m(ZnSO_{4})=\frac{\nu(Zn)\cdot N(ZnSO_{4})\cdot M(ZnSO_{4})}{N(Zn)}$

$M(ZnSO_{4})=1\cdot65.38+1\cdot32.06+4\cdot16=161.44\;amu$

$m(ZnSO_{4})=\frac{0.436\cdot 2\cdot 161.44}{2}\approx70.4\;g$

Answer: 70.4 g