The question requires us to balance a given reaction and then determine the excess reagent for this same reaction, given that 3.4 moles of Ca(NO3)2 and 2.4 moles of LiPO4 are used.
First, let's balance the equation so we can make the required calculations from it. A simple way to balance this reaction is to start with metals (Li and Ca) and consider the anions NO3 and PO4 instead of the isolated N, O and P atoms:
1) We write the given equation:
___Ca(NO3)2 + ____Li3PO4 --> ___LiNO3 + ____Ca3(PO4)2
2) We check the amount of Ca and Li on both sides of the reaction:
On the left side -> 1 Ca and 3 Li
On the right side -> 3 Ca and 1 Li
3) Then, we change the coefficients in order to have the same amount of Ca and Li on both sides:
3
4) Next, we check the amount of NO3 and PO4 anions:
Left side -> 3*2 NO3 = 6 NO3 and 1 PO4
Right side -> 3*1 NO3 = 3 and 2 PO4
4) We need to adjust some numbers in order to have the same amount of all considered species:
5) At last, we check again:
Left side -> 3 Ca, 6 NO3, 6 Li, 2 PO4
Right side -> 3 Ca, 6 NO3, 6 Li and 2 PO4
a) Then, the balanced reaction is:
1
b) Now that we know the stoichiometric coefficients, we can determine the excess reagent.
From the balanced reaction, we know that we need 3 moles of Ca(NO3)2 to react with 2 moles of Li3PO4. Then, we can write:
3 mol Ca(NO3)2 --------------- 2 mol Li3PO4
3.4 mol Ca(NO3)2 ------------ x
exexcesslimiting (more than 2.27 moles)
We can also determine the excess reactant from the following relation:
y ------------- 2.4 mol Li3PO4
In this case, solving for y, we have that we would need 3.6 moles of Ca(NO3)2 to react with 2.4 moles of Li3PO4, meaning that there is less than required of Ca(NO3)2, thus this is the limiting reactant while Li3PO4 is the excess reactant.