Answer:
Mass of Ba₃(PO₄)₂ = 0.0361 g
Explanation:
Given data:
Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )
Molarity of Ba(NO₃)₂ = 0.152 M
Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)
Molarity of K₃PO₄ = 0.604 M
Mass of Ba₃(PO₄)₂ produced = ?
Solution:
Chemical equation:
3Ba(NO₃)₂ + 2K₃PO₄ → Ba₃(PO₄)₂ + 6KNO₃
Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter
Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L
Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol
Number of moles of K₃PO₄ = Molarity × Volume in litter
Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L
Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol
Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .
Ba(NO₃)₂ : Ba₃(PO₄)₂
3 : 1
0.182 × 10⁻³ : 1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol
K₃PO₄ : Ba₃(PO₄)₂
2 : 1
2.537 × 10⁻³ : 1/2 × 2.537 × 10⁻³= 1.269 × 10⁻³ mol
The number of moles of Ba₃(PO₄)₂ produced by Ba(NO₃)₂ are less it will limiting reactant.
Mass of Ba₃(PO₄)₂ = moles × molar mass
Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol
Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g
Mass of Ba₃(PO₄)₂ = 0.0361 g
