Final answer:
In this double-displacement reaction, all of the calcium nitrate will be consumed, while none of the potassium phosphate will remain. The mass of calcium phosphate formed will be 364.3 g, and the mass of potassium nitrate formed will be 317.1 g.
Step-by-step explanation:
To determine the mass of each substance present after the reaction, we need to set up a balanced chemical equation and calculate the moles involved.
The balanced equation for the reaction between potassium phosphate (K3PO4) and calcium nitrate (Ca(NO3)2) is:
2K3PO4(aq) + 3Ca(NO3)2(aq) → Ca3(PO4)2(s) + 6KNO3(aq)
Now, we can use stoichiometry to determine the mass of each substance:
a) The initial mass of potassium phosphate is 500.0 g. Since it doesn't react completely, we need to determine the limiting reactant. We can calculate the moles of each reactant and compare their ratios in the balanced equation. The molar mass of K3PO4 is 212.27 g/mol, so we have:
Moles of K3PO4 = mass / molar mass = 500.0 g / 212.27 g/mol = 2.354 mol
Moles of Ca(NO3)2 = 2/3 * moles of K3PO4 = 1.569 mol
From the balanced equation, it's clear that 1 mol of K3PO4 reacts with 3 mol of Ca(NO3)2. Since there are fewer moles of Ca(NO3)2, it is the limiting reactant. Therefore, all of the Ca(NO3)2 will be consumed in the reaction.
b) The initial mass of calcium nitrate is 500.0 g. Since it is the limiting reactant, none of it will remain after the reaction.
c) The mass of calcium phosphate formed can be calculated by comparing the moles of K3PO4 and Ca3(PO4)2 in the balanced equation. From the mole ratios, we can determine that 2.354 mol of K3PO4 form 1.177 mol of Ca3(PO4)2. The molar mass of Ca3(PO4)2 is 310.18 g/mol, so:
Mass of Ca3(PO4)2 = moles * molar mass = 1.177 mol * 310.18 g/mol = 364.3 g
d) The mass of potassium nitrate formed can be calculated by multiplying the moles of Ca(NO3)2 consumed by the mole ratio in the balanced equation. From the balanced equation, we know that 3 mol of Ca(NO3)2 react to form 6 mol of KNO3. Since 1.569 mol of Ca(NO3)2 is consumed, we will form 3.138 mol of KNO3. The molar mass of KNO3 is 101.1 g/mol, so:
Mass of KNO3 = moles * molar mass = 3.138 mol * 101.1 g/mol = 317.1 g
Therefore, the mass of each substance present when the reaction stops is:
a) Potassium phosphate: 500.0 g - consumed in the reaction (none will remain)
b) Calcium nitrate: consumed in the reaction (none will remain)
c) Calcium phosphate: 364.3 g
d) Potassium nitrate: 317.1 g