Question

Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scientific notation formatting (ie, write 1.23e4 for 1.23x10^4). 9. b. Identify the mass of excess reagent remaining in the following system: lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: Li2O(s) + H2O(0) ► 2LIOH(s). In this system, 80.0kg of water is to be removed and 65.0kg of Li2O is available. Answer:

Answer 1

Answer:

The mass of excessive water (H₂O) is 40.815 kg

Explanation:

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.

Step 1: Calculating the number of moles of water to be removed

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

Step 2: Moles of Li₂O available

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

Step 3: Mass of excess reagent

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