Explanation:
What will the question be ?
Answer:
15.4 g of Zn₃(PO₄)₂ are produced
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
We shall find the molar mass first.
Ca5(PO4)3(OH) = (40 * 5) + 3 (31 + 4(16)) + 16 + 1 = 200 + 285 + 17 = 485 + 17 = 502.
Percent of Calcium = 200/ 502 * 100 = 39.8%
Percent of Phosphorus = 91/502 * 100 = 18.1%
Percent of Hydrogen = 1/502 * 100 = 0.19%
Percent of Oxygen = 100 - (39.8 + 18.1 + 0.19) = 41.91%
