A mixture is prepared by adding 28.6 mL of 0.169 M Na3PO4 to 39.7 mL of 0.255 M Ca(NO3)2. What mass of Ca3(PO4)2 will be formed?
You have:
.0286*.169 = 0.004833 moles of Na3PO4
.0397*.255 = 0.010124 moles of Ca(NO3)2
If the reaction is
2Na3PO4 + Ca(NO3)2 = 2NaNO3 + Ca(PO4)2
then you need 2 moles of Na for each mole of Ca.
Since there is extra Ca, you will wind up with
.09666 moles of Ca(PO4)2
Just multiply that by the mol mass of the compound.
To find the mass of Ca3(PO4)2 formed in the given mixture, we need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely consumed, thereby limiting the amount of product that can be formed.
To find the limiting reactant, we can compare the number of moles of each reactant. The balanced chemical equation for the reaction of Na3PO4 and Ca(NO3)2 is:
3 Na3PO4 + 2 Ca(NO3)2 → Ca3(PO4)2 + 6 NaNO3
From the equation, we can see that 3 moles of Na3PO4 react with 2 moles of Ca(NO3)2 to form 1 mole of Ca3(PO4)2.
1. Calculate the number of moles for each reactant:
Number of moles of Na3PO4 = volume (in L) × concentration (in mol/L)
= 28.6 mL × (1 L/1000 mL) × 0.169 mol/L
≈ 0.0048394 mol
Number of moles of Ca(NO3)2 = volume (in L) × concentration (in mol/L)
= 39.7 mL × (1 L/1000 mL) × 0.255 mol/L
≈ 0.0101715 mol
2. Determine the ratio of the moles of Na3PO4 to Ca(NO3)2:
Mole ratio of Na3PO4 to Ca(NO3)2 = (0.0048394 mol Na3PO4) / (0.0101715 mol Ca(NO3)2)
≈ 0.4754
Based on the mole ratio, it shows that there is an excess of Ca(NO3)2 compared to Na3PO4 (mole ratio < 1). Therefore, Na3PO4 is the limiting reactant.
3. Calculate the moles of Ca3(PO4)2 formed:
Since we have determined Na3PO4 as the limiting reactant, we can use its moles to calculate the moles of Ca3(PO4)2 formed.
Moles of Ca3(PO4)2 = (0.0048394 mol Na3PO4) × (1 mol Ca3(PO4)2 / 3 mol Na3PO4)
≈ 0.0016131 mol
4. Calculate the mass of Ca3(PO4)2 formed:
The molar mass of Ca3(PO4)2 can be found by adding up the atomic masses of the elements involved:
Molar mass of Ca3(PO4)2 = (3 × atomic mass of Ca) + (2 × atomic mass of P) + (8 × atomic mass of O)
= (3 × 40.08 g/mol) + (2 × 30.97 g/mol) + (8 × 16.00 g/mol)
≈ 310.18 g/mol
Mass of Ca3(PO4)2 formed = (0.0016131 mol Ca3(PO4)2) × (310.18 g/mol)
≈ 0.5 g
Therefore, approximately 0.5 grams of Ca3(PO4)2 will be formed in the given mixture.