Answer:
This means Ca(NO3)2 is the limiting reactant and Li3PO4 is the reactant in excess. LiNO3 and Ca3(PO4)2 are the products. Option C is correct.
Explanation:
Step 1: Data given
Number of moles Ca(NO3)2 = 3.4 moles
Number of moles Li3PO4 = 2.4 moles
Molar mass of Ca(NO3)2 = 164.09 g/mol
Molar mass of Li3PO4 =115.79 g/mol
Step 2: The balanced equation
3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2
Step 3: Calculate the limiting reactant
For 3 moles Ca(NO3)2 we need 2 moles Li3PO4 to produce 6 moles LiNO3 and 1 mol of Ca3(PO4)2
Ca(NO3)2 is the limiting reactant. It will completely be consumed. (3.4 moles). Li3PO4 is in excess. There will react 2/3 * 3.4 = 2.27 moles
There will remain 2.4 - 2.27 : 0.13 moles Li3PO4
This means Ca(NO3)2 is the limiting reactant and Li3PO4 is the reactant in excess. LiNO3 and Ca3(PO4)2 are the products. Option C is correct.