Question

Calcium carbonate reacts with phosphoric acid to produce calcium phosphate, carbon dioxide, and water. How many grams of phosphoric acid react with excess calcium carbonate to produce 3.74 grams of Ca3(PO4)2?

Answer 1

To find out how many grams of phosphoric acid react to produce 3.74 grams of calcium phosphate, stoichiometry is used to calculate that 2.35 grams of phosphoric acid are needed.

Step-by-step explanation:

The student asked how many grams of phosphoric acid are needed to react with excess calcium carbonate to produce 3.74 grams of calcium phosphate.

To answer this, we need to use the balanced chemical equation and stoichiometry:

Chemical Equation:

CaCO3(s) + 2 H3PO4(aq) → Ca3(PO4)2(s) + CO2(g) + 3 H2O(l)

First, we calculate the moles of Ca3(PO4)2 using its molar mass, and then we use the molar ratio from the balanced equation to find the moles of H3PO4. Finally, we convert these moles to grams using the molar mass of H3PO4.

Assuming we've already calculated the molar mass of Ca3(PO4)2 as 310.18 g/mol and the molar mass of H3PO4 as 98 g/mol:

1. Calculate moles of Ca3(PO4)2: (3.74 g) / (310.18 g/mol) = 0.012 mol
2. Find moles of H3PO4 needed using stoichiometry: (0.012 mol Ca3(PO4)2) × (2 mol H3PO4 / 1 mol Ca3(PO4)2) = 0.024 mol H3PO4
3. Convert moles of H3PO4 to grams: (0.024 mol) × (98 g/mol) = 2.35 g

Therefore, 2.35 grams of phosphoric acid are needed to react with excess calcium carbonate to produce 3.74 grams of calcium phosphate.

Answer 2

Mass Phosphoric acid needed = 2.36 g

Step-by-step explanation:

3CaCO₃ + 2H₃PO₄ => Ca₃(PO₄)₂ + 3CO₂ + 3H₂O

Given 3.74 g Ca₃(PO₄)₂ => (3.74g/310.2g·mol⁻¹) = 0.0121 mol Ca₃(PO₄)₂

=> 0.0121 mol Ca₃(PO₄)₂ requires 2(0.0121) mol H₃PO₄ = 0.0242 mol H₃PO₄

=> 0.0242 mol H₃PO₄ = (0.0242mol)(98g·mol⁻¹) = 2.36 g H₃PO₄