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Answer :
Final answer:
The maximum mass of water that could be produced by the chemical reaction between 36.3 g of sulfuric acid and 11.0 g of sodium hydroxide is 9.91 grams.
Explanation:
To calculate the maximum mass of water produced in the reaction between sulfuric acid and sodium hydroxide, we need to follow these steps:
- Convert the given masses of sulfuric acid and sodium hydroxide to moles.
- Determine the limiting reactant by comparing the moles of sulfuric acid and sodium hydroxide.
- Use the stoichiometry of the balanced equation to find the moles of water produced.
- Convert the moles of water to grams.
Step 1: Convert the given masses to moles:
Mass of sulfuric acid (H₂SO₄) = 36.3 g
Mass of sodium hydroxide (NaOH) = 11.0 g
Molar mass of H₂SO₄ = 98.09 g/mol
Molar mass of NaOH = 40.00 g/mol
Moles of H₂SO₄ = mass / molar mass = 36.3 g / 98.09 g/mol = 0.370 mol
Moles of NaOH = mass / molar mass = 11.0 g / 40.00 g/mol = 0.275 mol
Step 2: Determine the limiting reactant:
From the balanced equation, we can see that the stoichiometric ratio between H₂SO₄ and NaOH is 1:2. Therefore, for every 1 mole of H₂SO₄, we need 2 moles of NaOH.
Since we have 0.370 mol of H₂SO₄ and 0.275 mol of NaOH, the NaOH is the limiting reactant because we have less than half the amount needed to react with all the H₂SO₄.
Step 3: Use the stoichiometry to find the moles of water produced:
From the balanced equation, we know that for every 1 mole of H₂SO₄, 2 moles of water are produced.
Since NaOH is the limiting reactant, we can use its moles to determine the moles of water produced.
Moles of water = 2 moles of NaOH = 2 * 0.275 mol = 0.550 mol
Step 4: Convert the moles of water to grams:
Molar mass of H₂O = 18.02 g/mol
Mass of water = moles * molar mass = 0.550 mol * 18.02 g/mol = 9.91 g
Therefore, the maximum mass of water that could be produced by the chemical reaction is 9.91 grams.
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