Answer to Question #85771 in General Chemistry for Alexis

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