Consider the reaction.
Pb(SO4)2 + 2 Zn → 2 ZnSO4 +Pb
If 0.311 mol of zinc reacts with excess lead(IV) sulfate, how many grams of zinc sulfate will be produced in the reaction?

m = 50 g ZnSO4

Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.

0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn

= 0.311 moles ZnSO4

0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4

= 50 ZnSO4

mass ZnSO₄ = 50.2 gms (3 sig. figs.)

Explanation:

Pb(SO₄)₂(aq) + 2 Zn(s) => ZnSO₄(aq) + Pb(s)

=> Because coefficients of Zn and ZnSO₄ are the same, 0.311 mole Zn(s) yields the same 0.311 mol ZnSO₄(aq) .

=> Convert moles to grams by multiplying 0.311 moles ZnSO₄  by formula weight of ZnSO₄ (= 161.4 g/mol)

That is, grams ZnSO₄ = 0.311 mole ZnSO₄ x 161.4 g/mol = 50.1954 g ≅ 50.2 g (3 sig. figs.)

answer: h+ its less than oh

explanation: