Answer:
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Explanation:
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
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Zn+2HCl ----> 2ZnCl2 + H2
For 2.50 g of Zn
Mass per mol = 2.50/molar mass of Zn = 2.50/65.38 = 0.0382 g/mol
There are two moles of ZnCl2 and total mass = 2*0.0382*molar mass of ZnCl2 = 2*0.0382*136.286 = 10.42 g
For 2 g of HCl
Mass per mol = 2/2*molar mass of HCl = 2/ (2*36.46) = 0.0274 g/mol
For the two moles of ZnCl2, mass produced = 2*0.0274*136.286 = 7.48 g
It can be noted that 2 g of HCl produced less amount of ZnCl and thus it is the limiting reagent.