What is the mole fraction of water in 200g of 95 ethanol?
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<h3>Answer:</h3>
The mass of excessive water (H₂O) is 40.815 kg
<h3>Explanation:</h3>The Equation for the reaction is;
Li₂O(s) + H₂O(l) → 2LiOH(s)
From the question;
Mass of water removed is 80.0 kg
Mass of available Li₂O is 65.0 kg
We are required to calculate the mass of excessive reagent.
<h3>Step 1: Calculating the number of moles of water to be removed</h3>Moles = Mass ÷ Molar mass
Molar mass of water = 18.02 g/mol
Mass of water = 80 kg (but 1000 g = 1kg)
= 80,000 g
Therefore;
= 4.44 × 10³ moles
<h3>Step 2: Moles of Li₂O available </h3>Moles = mass ÷ molar mass
Mass of Li₂O available = 65.0 kg or 65,000 g
Molar mass Li₂O = 29.88 g/mol
Moles of Li₂O = 65,000 g ÷ 29.88 g/mol
= 2.175 × 10³ moles Li₂O
<h3>Step 3: Mass of excess reagent </h3>From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)
1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH
The ratio of Li₂O to H₂O is 1:1
- Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
- However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
- This means, Li₂O is the limiting reactant while water is the excessive reagent.
Therefore:
Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles
= 2.265 × 10³ moles
Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol
= 4.0815 × 10⁴ g or
= 40.815 kg
Thus, the mass of excessive water is 40.815 kg
When base is added in order to increase the pH of a solution, it results in MOH salt <u>precipitating first.</u>
Equation :
MOH (s) ⇄ M⁺ (aq) ₊ OH⁻(aq)
K = [M⁺] [OH⁻]
K = 1.39×10⁻¹² and [M⁺] = 0.001M
∴ 1.39×10⁻¹² = (0.001M) [OH⁻]
[OH⁻] = 1.004×10⁻¹⁹M
pOH = -log [OH⁻] = 11.9
pH = 14 - pOH
= 14 - 11.9
<u>pH = 2.1</u>
To learn more about solubility equilibrium ;
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