The question requires us to calculate the amount, in grams, of calcium phosphate that would be produced, given the reaction and the initial amount of reactants.
The following information is provided by the question:
- the chemical reaction already balanced:
3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2
- the initial amount of both reactants:
number of moles of Ca(NO3)2 = 3.40 mol
number of moles of Li3PO4 = 2.40 mol
To answer this question, we need to go through the following steps:
1) check if the reaction is balanced;
2) check if there is a limiting reactant and which it is, since the amount of both reactants was provided;
3) calculate the amount of Ca3(PO4)2 that would be produced from the limiting reactant;
4) calculate the molar mass of Ca3(PO4)2 and use to obtain the mass of this compound produced.
Next, we'll solve the problem following the steps:
1) The reaction provided is already balanced (there is the same amount of atoms of each element on both sides of the equation)
2) To check if there is a limiting reactant, we can use the stoichiometric relation provided by the reaction:
3 mol of Ca(NO3)2 --------------- 2 mol of Li3PO4
Given the stoichiometric relation, we use it with the number of moles provided for each reactant and obtain what number of moles would be theoretically necessary:
3 mol Ca(NO3)2 --------------- 2 mol Li3PO4
3.40 mol Ca(NO3)2 ---------- x
Solving for x, we have that 2.27 moles of Li3PO4 would be necessary to react with 3.40 moles of Ca(NO3)2.
We can use this same logic for Li3PO4:
3 mol Ca(NO3)2 --------------- 2 mol Li3PO4
y ------------------------------------ 2.40 mol Li3PO4
Solving for y, we have that 3.6 moles of Ca(NO3)2 would be necessary to react with 2.40 moles of Li3PO4.
From the calculations above, we can notice that Ca(NO3)2 is the limiting reactant in this case, and we should use it for the next steps.
3) Next, we use the number of moles for the limiting reactant (3.40 moles of Ca(NO3)2) and the stoichiometric relation given by the reaction to calculate how many moles of Ca3(PO4)2 would be produced:
3 mol Ca(NO3)2 --------------- 1 mol Ca3(PO4)2
3.40 mol Ca(NO3)2 ---------- z
Solving for z, we have that 1.13 moles of Ca3(PO4)2 would be obtained.
4) The last step is convert the number of moles obtained for Ca3(PO4)2 in its correspondent mass. For this calculation, we'll need the molar mass of this compound.
The atomic masses of Ca, P and O are 40.08 u, 30.97 u and 15.99 u, respectively. Considering the amount of atoms of each of this element in the molecule Ca3(PO4)2:
molar mass Ca3(PO4)2 = (3 * 40.08) + (2 * 30.97) + (8 * 15.99) = 310.1 g/mol
Now, with the molar mass of Ca3(PO4)2 and the calculated number of moles of this compound (1.13 moles), we can calculate the mass that would be produced:
1 mol Ca3(PO4)2 --------------- 310.1 g/mol
1.13 mol Ca3(PO4)2 ----------- w
Solving for w, we have that 350.4 grams of Ca3(PO4)2 would be obtained.
Therefore, the best option to answer the question would be the last one, 351 g.