Question

To ensure that the air in the shuttle remains free of excess CO2, engineers test the air-purification system. they combine 1.000 × 103 g LiOH with 8.80 × 1 02 g CO2. The reaction produces 3.25 × 102 g H2O. What is the limiting reactants in this test reaction? What is the theoretical yield of this test reaction?

Answer 1

The

balanced reaction is:

2LiOH + CO2 = Li2CO3 + H2O

We

are given the amount of the reactants to be reacted. This will be the starting point of our

calculations. To determine the limiting reactant, we convert the amounts from grams to moles.

1.000 × 10^3 g LiOH  (1 mol / 25.95 g) = 38.54 mol LiOH

8.80 × 10^2 g CO2 ( 1mol / 44.01 g) = 20.00 mol CO2

From the balanced reaction, the molar ratio of the reactants is 2:1. This means that every two mole of lithium hydroxide, 1 mole of carbon dioxide is needed. It is clear that, from the given amounts, CO2 is the limiting reactant.

Answer 2

Answer: The limiting reagent is carbon dioxide and theoretical yield of the reaction is 360 g.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For LiOH:

Given mass of LiOH = $1.00\times 10^3g$

Molar mass of LiOH = 23.95 g/mol

Putting values in equation 1, we get:

$\text{Moles of LiOH}=\frac{1.00\times 10^3g}{23.95g/mol}=41.75mol$

• For carbon dioxide:

Given mass of carbon dioxide = $8.8\times 10^2g$

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide}=\frac{8.8\times 10^2g}{44g/mol}=20mol$

• The chemical equation for the reaction of lithium hydroxide and carbon dioxide gas follows:

$2LiOH(s)+CO_2(g)\rightarrow Li_2CO_3(s)+H_2O(l)$

By Stoichiometry of the reaction:

1 moles of carbon dioxide reacts with 2 moles of lithium hydroxide.

So, 20 moles of carbon dioxide will react with = $\frac{2}{1}\times 20=40mol$ of lithium hydroxide.

As, given amount of lithium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, carbon dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of carbon dioxide produces 1 mole of water

So, 20 moles of carbon dioxide will produce = $\frac{1}{1}\times 20=20moles$ of water

• Now, calculating the theoretical yield of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 20 moles

Putting values in equation 1, we get:

$20mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=360g$

Hence, the limiting reagent is carbon dioxide and theoretical yield of the reaction is 360 g.