Calcium carbonate reacts with phosphoric acid to produce calcium phosphate, carbon dioxide, and water.
3 CaCO3(s) + 2 H3PO4(aq) Ca3(PO4)2(aq) + 3 CO2(g) + 3 H2O(l)
How many grams of phosphoric acid react with excess calcium carbonate to produce 3.47 g Ca3(PO4)2?
Here is a worked example of a stoichiometry problem similar to yours. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the grams of phosphoric acid required, we need to use the molar ratios from the balanced equation.
From the balanced equation, we can see that:
3 moles of calcium carbonate react with 2 moles of phosphoric acid to produce 1 mole of calcium phosphate (Ca3(PO4)2).
Given:
Mass of Ca3(PO4)2 = 3.47 g
To find the moles of Ca3(PO4)2, we will use its molar mass:
Molar mass of Ca3(PO4)2 = (40.08 g/mol * 3) + (30.97 g/mol * 2) + (16.00 g/mol * 8) = 310.18 g/mol
Moles of Ca3(PO4)2 = Mass / Molar mass = 3.47 g / 310.18 g/mol = 0.0112 mol
Now, using the molar ratios from the balanced equation, we can determine the moles of phosphoric acid required.
From the balanced equation:
3 moles of Ca3(PO4)2 corresponds to 2 moles of H3PO4
Moles of H3PO4 = Moles of Ca3(PO4)2 * (2 moles H3PO4 / 3 moles Ca3(PO4)2)
= 0.0112 mol * (2/3)
= 0.00747 mol
Lastly, we can calculate the mass of phosphoric acid using its molar mass.
Molar mass of H3PO4 = (1.01 g/mol * 3) + (16.00 g/mol) + (1.01 g/mol * 4) = 97.99 g/mol
Mass of H3PO4 = Moles of H3PO4 * Molar mass
= 0.00747 mol * 97.99 g/mol
= 0.734 g
Therefore, approximately 0.734 grams of phosphoric acid would react with excess calcium carbonate to produce 3.47 g of Ca3(PO4)2.
To determine the mass of phosphoric acid required to produce 3.47 g of calcium phosphate (Ca3(PO4)2), you can use the balanced chemical equation and stoichiometry.
1. Determine the molar mass of calcium phosphate (Ca3(PO4)2):
- Calcium (Ca): 3 atoms x atomic mass of calcium (40.08 g/mol) = 120.24 g/mol
- Phosphorus (P): 1 atom x atomic mass of phosphorus (30.97 g/mol) = 30.97 g/mol
- Oxygen (O): 8 atoms x atomic mass of oxygen (16.00 g/mol) = 128.00 g/mol
- Total molar mass = 120.24 g/mol + 30.97 g/mol + 128.00 g/mol = 279.21 g/mol
2. Use the stoichiometry of the balanced equation to determine the mole ratio between calcium phosphate and phosphoric acid:
- From the balanced equation, 2 moles of phosphoric acid react with 1 mole of calcium phosphate.
3. Convert the mass of calcium phosphate to moles:
- Moles of calcium phosphate = mass (g) / molar mass (g/mol)
- Moles of calcium phosphate = 3.47 g / 279.21 g/mol ≈ 0.01244 mol
4. Use the mole ratio from step 2 to determine the moles of phosphoric acid:
- Moles of phosphoric acid = 2 moles of phosphoric acid * moles of calcium phosphate
= 2 * 0.01244 mol ≈ 0.02488 mol
5. Finally, convert the moles of phosphoric acid to grams:
- Mass of phosphoric acid = moles of phosphoric acid * molar mass of phosphoric acid
= 0.02488 mol * molar mass of phosphoric acid
Note: The molar mass of phosphoric acid (H3PO4) is:
- Hydrogen (H): 3 atoms x atomic mass of hydrogen (1.01 g/mol) = 3.03 g/mol
- Phosphorus (P): 1 atom x atomic mass of phosphorus (30.97 g/mol) = 30.97 g/mol
- Oxygen (O): 4 atoms x atomic mass of oxygen (16.00 g/mol) = 64.00 g/mol
- Total molar mass = 3.03 g/mol + 30.97 g/mol + 64.00 g/mol = 98.00 g/mol
- Mass of phosphoric acid = 0.02488 mol * 98.00 g/mol ≈ 2.44 g
Therefore, approximately 2.44 grams of phosphoric acid will react with excess calcium carbonate to produce 3.47 grams of calcium phosphate.