A solution that has a relatively large quantity of solute dissolved in the solvent is:? and please explain the answer
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2. 15.65 grams of KCl is formed if 25.0 g of Potassium Chlorate, KClO3, decompose.
3. 2.17 moles of H2 gas produced when 100.0 grams of Na is added to the reaction.
4. 7.5 moles of Hydrogen, H2, are needed to react with 2.5 moles of Nitrogen, N2.
Explanation:
2. The balanced equation for the chemical reaction is:
2KClO3 ⇒ 2KCl + 3O2
Data given:
mass of KClO3 = 25 grams, atomic mass of KClO3 = 122.55 grams/mole
KCl produced =? atomic mass of KCl = 74.55
number of moles =
=
number of moles = 0.20 moles of KCO3
2 moles of KClO3 decomposes to give 2KCl
0.21 moles of KClO3 decomposes to give x moles of KCl
=
x = 0.21 moles of KCl
mass of KCl = 0.21 x 74.55
= 15.65 grams of KCl is formed.
3. data given:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
mass of Na added = 100 grams, atomic mass of Na = 22.98 grams/mole
moles of H2 =?
Number of moles of Na =
=
number of moles of Na = 4.35 moles
2 moles of Na gives 1 mole of H2
4.35 moles of Na will give x moles
=
2x = 4.35
x = 2.17 moles of hydroden gas is produced.
4. Data given:
the balanced chemical equation:
N2 + 3H2 → 2NH3
Number of moles of N2 2.5
from the reaction
1 mole of N2 reacts with 3 moles of H2
The molar ratio of the reactant is 1:3
so, =
x = 7.5 moles of H2 will react with 2.5 moles of N2.
<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