Answer:
15.4 g of Zn₃(PO₄)₂ are produced
Step-by-step explanation:
Given data:
Mass of zinc phosphate formed = ?
Volume of zinc nitrate = 48.1 mL (0.05 L)
Molarity of zinc nitrate = 2.18 M
Solution:
Chemical equation:
3Zn(NO₃)₂ + 2K₃PO₄ → Zn₃(PO₄)₂ + 6KNO₃
Moles of zinc nitrate:
Molarity = number of moles / volume in litter
Number of moles = 2.18 M × 0.05 L
Number of moles = 0.109 mol
Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:
Zn(NO₃)₂ : Zn₃(PO₄)₂
3 : 1
0.109 : 1/3×0.109 = 0.04 mol
0.04 moles of Zn₃(PO₄)₂ are produced.
Mass of Zn₃(PO₄)₂:
Mass = number of moles × molar mass
Mass = 0.04 mol × 386.1 g/mol
Mass = 15.4 g