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CO2 + 2LiOH = Li2CO3 + H2O

To ensure that the air in the shuttle remains free of excess CO2, engineers test the air-purification system.

They combine 1.000 × 10^3 g LiOH with 8.80 × 10^2 g CO2. The reaction produces 3.25 × 10^2 g H2O.

What is the theoretical yield of this test reaction? Record answer to the nearest whole number.

_g H2O

Verified Answer :

Answer : The theoretical yield of water in this reaction is 360 g.

Solution : Given,

Mass of \(LiOH=1\times 10^3g=1000g\)

Mass of \(CO_2=8.80\times 10^2g=880g\)

Mass of \(H_2O=3.25\times 10^2g=325g\)

Molar mass of \(CO_2\) = 44 g/mole

Molar mass of \(LiOH\) = 24 g/mole

Molar mass of \(H_2O\) = 18 g/mole

First we have to calculate the moles of \(CO_2\) and \(LiOH\).

Moles of \(CO_2\) = \(\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}= \frac{880g}{44g/mole}=20moles\)

Moles of \(LiOH\) = \(\frac{\text{ Mass of }LiOH}{\text{ Molar mass of }LiOH}= \frac{1000g}{24g/mole}=42moles\)

The given balanced chemical reaction is,

\(CO_2+2LiOH\rightarrow Li_2CO_3+H_2O\)

From the given reaction, we conclude that

1 mole of \(CO_2\) react with 2 mole of \(LiOH\)

20 moles of \(CO_2\) requires \(20\times2=40moles\) of \(LiOH\)

But the actual moles of \(LiOH\) = 42 moles

Excess moles of \(LiOH\) = 42 - 40 = 2 moles

So, \(CO_2\) is the limiting reagent.

Now we have to calculate the mass of \(H_2O\).

From the reaction, we conclude that

1 mole of \(CO_2\) gives 1 mole of \(H_2O\)

20 moles of \(CO_2\) gives 20 moles of \(H_2O\)

Mass of \(H_2O\) = Moles of \(H_2O\) × Molar mass of  \(H_2O\)

Mass of \(H_2O\) = 20 moles × 18 g/mole = 360 g

The experimental yield of \(H_2O\) = 325 g

Therefore, the theoretical yield of \(H_2O\) = 360 g

Answer:

Part 1: limiting reactant is CO2

Part 2: theoretical yield is 360

Part 3: percent yield is 90.3

Explanation:

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