To identify the limiting reactant, excess reactant, and theoretical yield, we first need to determine the amount of each reactant in moles.
Using the molar masses of Ca and P:
Number of moles of Ca = 17.0 g / 40.08 g/mol = 0.424 mol
Number of moles of P = 18.0 g / 30.97 g/mol = 0.581 mol
Next, we need to determine the stoichiometric ratio of the reactants. From the balanced chemical equation, we see that the ratio of Ca to P is 3:2.
3 Ca + 2 P → Ca3P2
To use the stoichiometric ratio to determine the limiting reactant, we need to compare the actual ratio of the reactants to the stoichiometric ratio.
Actual ratio of Ca to P = (0.424 mol Ca) / (0.581 mol P) ≈ 0.73
Stoichiometric ratio of Ca to P = 3/2 = 1.5
Since the actual ratio is greater than the stoichiometric ratio, Ca is the excess reactant and P is the limiting reactant.
To find the theoretical yield of Ca3P2, we need to use the stoichiometric ratio to determine how many moles of Ca3P2 can be produced from the limiting reactant (P).
From the balanced chemical equation, we see that 2 moles of P react with 3 moles of Ca to produce 1 mole of Ca3P2.
So, the number of moles of Ca3P2 that can be produced from 0.581 mol of P is:
(0.581 mol P) × (1 mol Ca3P2 / 2 mol P) = 0.2905 mol Ca3P2
Therefore, the theoretical yield of Ca3P2 is 0.2905 mol.