Ca3(PO4)2 + 3 H2SO4 -> 3 CaSO4 + 2 H3PO4
What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?
52.8
To solve this problem, we need to follow these steps:
1. Write the balanced chemical equation:
Ca3(PO4)2 + 3 H2SO4 -> 3 CaSO4 + 2 H3PO4
2. Convert the given masses of reactants into moles by using their molar masses:
Mass of Ca3(PO4)2 = 1.0 kg
Molar mass of Ca3(PO4)2 = (3 x molar mass of Ca) + (2 x molar mass of P) + (8 x molar mass of O)
= (3 x 40.08 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)
= 310.18 g/mol
Number of moles of Ca3(PO4)2 = Mass / Molar mass
= 1000 g / 310.18 g/mol
= 3.223 mol
Mass of H2SO4 = 1.0 kg x 0.98 (as it is 98% by mass)
= 0.98 kg
Molar mass of H2SO4 = (2 x molar mass of H) + (1 x molar mass of S) + (4 x molar mass of O)
= (2 x 1.01 g/mol) + (1 x 32.07 g/mol) + (4 x 16.00 g/mol)
= 98.09 g/mol
Number of moles of H2SO4 = Mass / Molar mass
= 980 g / 98.09 g/mol
= 9.978 mol
3. Determine the limiting reactant:
To find the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.
From the balanced equation, we can see that 1 mol of Ca3(PO4)2 reacts with 3 mol of H2SO4, so the mole ratio is:
3 mol of H2SO4 / 1 mol of Ca3(PO4)2
Since we have 9.978 mol of H2SO4, we can calculate the equivalent moles of Ca3(PO4)2 that would be required:
9.978 mol H2SO4 x (1 mol Ca3(PO4)2 / 3 mol H2SO4) = 3.326 mol Ca3(PO4)2
As we have 3.223 mol of Ca3(PO4)2, it is the limiting reactant because it has fewer moles than required.
4. Calculate the masses of the products:
Using the mole ratio from the balanced equation, we can calculate the number of moles of the other products.
From the balanced equation, we can see that 3 mol of CaSO4 is produced from 1 mol of Ca3(PO4)2, so the mole ratio is:
3 mol CaSO4 / 1 mol Ca3(PO4)2
Number of moles of CaSO4 = 3.223 mol of Ca3(PO4)2 x (3 mol CaSO4 / 1 mol Ca3(PO4)2) = 9.669 mol CaSO4
Mass of CaSO4 = Number of moles x Molar mass
= 9.669 mol x (1 x 40.08 g/mol + 1 x 32.07 g/mol + 4 x 16.00 g/mol)
= 981.53 g
From the balanced equation, we can see that 2 mol of H3PO4 is produced from 1 mol of Ca3(PO4)2, so the mole ratio is:
2 mol H3PO4 / 1 mol Ca3(PO4)2
Number of moles of H3PO4 = 3.223 mol of Ca3(PO4)2 x (2 mol H3PO4 / 1 mol Ca3(PO4)2) = 6.446 mol H3PO4
Mass of H3PO4 = Number of moles x Molar mass
= 6.446 mol x (3 x 1.01 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol)
= 429.61 g
Therefore, the masses of calcium sulfate and phosphoric acid that can be produced are approximately 981.53 g and 429.61 g, respectively.
This is a limiting reagent problem.
1. You have the equation.
2a. Convert 1 kg calcium phosphate to moles. moles = grams/molar mass.
2b. Convert 1 kg H2SO4 to moles. Same process BUT you are starting with 1,000 x 0.98 grams and not 1 kg.
3a. Using the coefficients in the balanced equation, convert moles Ca3(PO4)2 to moles CaSO4.
3b. Use the same process to convert moles H2SO4 to moles CaSO4.
3c. It is quite likely that the answers for 3a and 3b will not be the same which means one of them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reactant producing the smaller value is the limiting reagent.
4. Now convert moles CaSO4 to grams. g = moles x molar mass.
5. Repeat steps 2a, 2b, 3a, 3b, 3c, and 4 for H3PO4. A shorter way is to recognize the limiting reagent from the first part of the problem with CaSO4 and applying that information to the determination of the grams H3PO4.