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