How many grams of calcium phosphate are theoretically produced if we start with 10.3 moles of Ca(NO3)2 and 7.6 moles of Li3PO4?
Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2
Here is a worked example of a limiting reagent problem.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
How many grams of calcium phosphate are theoretically produced if we start with 10.3 moles of Ca(NO3)2 and 7.6 moles of Li3PO4?
Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2
To answer this question, we need to determine the stoichiometry of the balanced chemical equation for the reaction between calcium nitrate (Ca(NO3)2) and lithium phosphate (Li3PO4) to produce calcium phosphate.
The balanced chemical equation for this reaction is:
3 Ca(NO3)2 + 2 Li3PO4 → Ca3(PO4)2 + 6 LiNO3
From this equation, we can see that 3 moles of Ca(NO3)2 react with 2 moles of Li3PO4 to produce 1 mole of Ca3(PO4)2.
Given that you have 10.3 moles of Ca(NO3)2 and 7.6 moles of Li3PO4, we need to determine which reactant is the limiting reactant, as it will determine the maximum amount of calcium phosphate produced.
To find the limiting reactant, we compare the mole ratios of the two reactants to the stoichiometry of the balanced equation.
For Ca(NO3)2:
Mole ratio: 3 moles Ca(NO3)2 : 1 mole Ca3(PO4)2
For Li3PO4:
Mole ratio: 2 moles Li3PO4 : 1 mole Ca3(PO4)2
Now, using the mole ratios, we can calculate the moles of calcium phosphate that can be produced from each reactant:
From Ca(NO3)2:
10.3 moles Ca(NO3)2 × (1 mole Ca3(PO4)2 / 3 moles Ca(NO3)2) = 3.43 moles Ca3(PO4)2
From Li3PO4:
7.6 moles Li3PO4 × (1 mole Ca3(PO4)2 / 2 moles Li3PO4) = 3.8 moles Ca3(PO4)2
Based on the calculations, we can see that the limiting reactant is Ca(NO3)2 because it produces a lower amount of Ca3(PO4)2 compared to Li3PO4.
Now, to find the theoretical mass of calcium phosphate produced, we need to use the molar mass of Ca3(PO4)2, which is calculated as:
Molar mass of Ca3(PO4)2 = (3 × Atomic mass of Ca) + (2 × Atomic mass of P) + (8 × Atomic mass of O)
Molar mass of Ca3(PO4)2 = (3 × 40.08 g/mol) + (2 × 30.97 g/mol) + (8 × 16.00 g/mol) = 310.18 g/mol
Finally, we can calculate the mass of calcium phosphate using the moles of Ca3(PO4)2 and its molar mass:
Mass of Ca3(PO4)2 = moles of Ca3(PO4)2 × molar mass of Ca3(PO4)2
Mass of Ca3(PO4)2 = 3.43 moles × 310.18 g/mol = 1063.5 grams
Therefore, theoretically, 1063.5 grams of calcium phosphate can be produced from 10.3 moles of Ca(NO3)2 and 7.6 moles of Li3PO4.