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3 years ago

I need yall help with this one and i can make yall brilientist if u get it right plus yall get 50 points

Chemistry

2 answers:

mixas84 [53]3 years ago

6 0

Answer:

Explanation:

amid [387]3 years ago

4 0

Answer:

C. and B. im pretty sure hope it helps

Explanation:

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which of the following is a physical change? 1.burning paper 2.grinding wheat 3.electrolysis of water 4.cooking rice

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2 grinding of wheat is just a physical change

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3 years ago

What was one main point of daltons atomic theory

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4 years ago

1. How will you differentiate physical from chemical change in terms of the

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3 years ago

For the closed system below: Cu(s) + 2Ag+(aq) 2Ag(s) + Cu2+(aq) How would you know if the system is at equilibrium?

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3 years ago

Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti

djverab [1.8K]

<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;

$Moles of water = \frac{80,000g}{18.02 g/mol}$

= 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

7 0

3 years ago