If 2.400 Moles of calcium nitrate reacts with 393.70g lithium phosphate, how much precipitate in grams will be produced?

Method 1 :

Molar mass of Li₃PO₄ = (6.94×3 + 30.97 + 16.00×4) g/ mol = 115.79 g/mol
Molar mass of Ca₃(PO₄)₂ = (40.08×3 + 30.97×2 + 16.00×8) g/mol = 310.18 g/mol

3Ca(NO₃)₂(aq) + 2Li₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6LiNO₃(s)
Mole ratio Ca(NO₃)₂ : Li₃PO₄ = 3 : 2

Initial no. of moles of Ca(NO₃)₂ = 2.400 mol
Initial no. of moles of Li₃PO₄ = (393.70 g) / (115.79 g/mol) = 3.400 mol

When all the 2.400 mol of Ca(NO₃)₂ completely reacts :
Number of moles of Li₃PO₄ needed = (2.400 mol) × (2/3) = 1.600 mol < 3.400 mol
Hence, Li₃PO₄ is in excess, and Ca(NO₃)₂ is the limiting reactant (completely reacts).

According to the equation, mole ratio Ca(NO₃)₂ : Ca₃(PO₄)₂ = 3 : 1
No. of moles of Ca₃(PO₄)₂ ppt produced = (2.400 mol) × (1/3) = 0.8000 mol
Mass of Ca₃(PO₄)₂ ppt produced = (310.18 g/mol) × (0.8000 mol) = 248.1 g


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Method 2 :
Molar mass of Li₃PO₄ = (6.94×3 + 30.97 + 16.00×4) g/ mol = 115.79 g/mol
Molar mass of Ca₃(PO₄)₂ = (40.08×3 + 30.97×2 + 16.00×8) g/mol = 310.18 g/mol

3Ca(NO₃)₂(aq) + 2Li₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6LiNO₃(s)

Mole ratio Ca(NO₃)₂ : Ca₃(PO₄)₂ = 3 : 1
Initial no. of moles of Ca(NO₃)₂ = 2.400 mol
If Ca(NO₃)₂ completely reacts, no. of moles of Ca₃(PO₄)₂ ppt produced = (2.400 mol) × (1/3) = 0.800 mol

Mole ratio Li₃PO₄ : Ca₃(PO₄)₂ = 2 : 1
Initial no. of moles of (393.70 g) / (115.79 g/mol) = 3.400 mol
If Li₃PO₄ completely reacts, no. of moles of Ca₃(PO₄)₂ ppt produced = (3.400 mol) × (1/2) = 1.700 mol

The limiting reactant produces a smaller amount of product.
No. of moles of Ca₃(PO₄)₂ ppt produced = (2.400 mol) × (1/3) = 0.800 mol
Mass of Ca₃(PO₄)₂ ppt produced = (310.18 g/mol) × (0.8000 mol) = 248.1 g

Start by writing a balanced equation for the reaction:
3Ca(NO3)2 (aq) + 2Li3PO4(aq) → Ca3(PO4)2 (s) + 6LiNO3(aq)
The Ca3(PO4)2 is a precipitate
3mol Ca(NO3)2 react with 2 mol Li3PO4 to produce 1mol Ca3(PO4)2
You have 2.4 mol Ca(NO3)2
24.mol Ca(NO3)2 will react with 2/3*2.4 = 1.6 mol Li3PO4.
Molar mass of Li3PO4 = 115.79g/molM
Mol Li3PO4 in 393.70g = 393.70/115.79 = 3.40mol
You have excess Li3PO4 The Ca(NO3)2 is limiting
Frome the equation, 3mol Ca(NO3)2 will produce 1 mol Ca3(PO4)2
2.4 mol Ca(NO3)2 will produce 1/3*2.4 = 0.8 mol Ca3(PO4)2
Molar mass Ca3(PO4)2 = 310.18 g/mol
Mass of 0.8g = 248.144g Ca3(PO4)2 precipitates.
Answer to 4 significant figures: 248.1g precipitate formed.