HELP ME PLEASE!!!!! I NEED TO GET THIS RIGHT I CANT FAIL
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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
Answer : The vapor pressure of solution is 23.67 mmHg.
Solution:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
where,
= vapor pressure of pure solvent (water) = 23.76 mmHg
= vapor pressure of solution= ?
= mass of solute (sucrose) = 12.25 g
= mass of solvent (water) = 176.3 g
= molar mass of solvent (water) = 18.02 g/mole
= molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in this formula ,we get the vapor pressure of the solution.
Therefore, the vapor pressure of solution is 23.67 mmHg.
Answer:
Explanation:
An empirical formula shows the smallest whole-number ratio of the atoms in a compound.
So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.
<h3>1. Divide </h3>The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.
There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300
- Sulfur: 0.300/0.300= 1
- Oxygen: 0.900/0.300= 3
The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.
- S₁O₃
This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.
- SO₃