Given the following reaction: Ca(OH)2 + H2SO4 → CaSO4 + 2H2O If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4, a) Determine the limiting reagent. b) Determine the number of moles of H2O produced. c) Determine the number of grams of CaSO4 produced. d) Determine the number of grams of excess reagent left.
The Correct Answer and Explanation is:
To solve this stoichiometry problem, we will follow these steps:
Balanced Chemical Equation:
Ca(OH)2+H2SO4→CaSO4+2H2O\text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O}Ca(OH)2+H2SO4→CaSO4+2H2O
This equation tells us that 1 mole of calcium hydroxide reacts with 1 mole of sulfuric acid to produce 1 mole of calcium sulfate and 2 moles of water.
Step 1: Molar Masses
- Ca(OH)₂ = 40.08 + (16.00 + 1.008) × 2 = 74.096 g/mol
- H₂SO₄ = 2(1.008) + 32.06 + 4(16.00) = 98.076 g/mol
- CaSO₄ = 40.08 + 32.06 + 4(16.00) = 136.14 g/mol
- H₂O = 2(1.008) + 16.00 = 18.016 g/mol
Step 2: Convert Mass to Moles
- Moles of Ca(OH)₂ = 14.82 g ÷ 74.096 g/mol = 0.2000 mol
- Moles of H₂SO₄ = 16.35 g ÷ 98.076 g/mol = 0.1667 mol
Step 3: Identify the Limiting Reagent
From the balanced equation, the molar ratio is 1:1.
Compare the actual moles available:
- Ca(OH)₂: 0.2000 mol
- H₂SO₄: 0.1667 mol
Since H₂SO₄ has fewer moles than Ca(OH)₂, H₂SO₄ is the limiting reagent.
Step 4: Calculate Moles of H₂O Produced
From the equation: 1 mol H₂SO₄ → 2 mol H₂O
So, 0.1667 mol H₂SO₄ → 0.1667 × 2 = 0.3334 mol H₂O
Step 5: Calculate Grams of CaSO₄ Produced
From the equation: 1 mol H₂SO₄ → 1 mol CaSO₄
So, 0.1667 mol CaSO₄ × 136.14 g/mol = 22.69 g CaSO₄
Step 6: Calculate Excess Reagent Left (Ca(OH)₂)
Used Ca(OH)₂ = 0.1667 mol
Initial = 0.2000 mol
Remaining = 0.2000 – 0.1667 = 0.0333 mol
Mass = 0.0333 mol × 74.096 g/mol = 2.47 g Ca(OH)₂ left
Final Answers:
a) Limiting reagent: H₂SO₄
b) Moles of H₂O produced: 0.3334 mol
c) Grams of CaSO₄ produced: 22.69 g
d) Grams of excess Ca(OH)₂ left: 2.47 g
Explanation
This chemical reaction between calcium hydroxide and sulfuric acid is a typical acid-base reaction. To determine the limiting reagent, we first need to calculate the moles of each reactant by dividing their given masses by their molar masses. The balanced equation shows a 1:1 mole ratio between the two reactants, so we directly compare the available moles. The reactant with fewer moles is the limiting reagent because it will be completely consumed first, stopping the reaction.
In this case, sulfuric acid is the limiting reagent since we only have 0.1667 moles compared to 0.2000 moles of calcium hydroxide. Therefore, all calculations for product formation are based on the amount of sulfuric acid available.
Using stoichiometry, we find that every mole of sulfuric acid produces two moles of water. Thus, 0.1667 moles of sulfuric acid yield 0.3334 moles of water. To find the mass of calcium sulfate formed, we multiply the number of moles of CaSO₄ (equal to the moles of H₂SO₄ used) by its molar mass.
Lastly, to determine how much calcium hydroxide remains, we subtract the amount that reacted from the initial amount and convert the remaining moles back into grams. This provides the mass of the excess reagent left over after the reaction.
By following this structured approach, we accurately determine the limiting reagent, the amount of water and calcium sulfate produced, and how much excess calcium hydroxide remains.
