Question

if 14.0 grams of co₂ are removed from the lunar module air, how many molecules of li₂co₃ are produced by the reaction? show all of your work. you must be able to prove your answer. 1.91 x 10^23 molecules 3.52 x 10^23 molecules 6.02 x 10^23 molecules 1.20 x 10^24 molecules

Explanation:

Step1: Calculate moles of \( \text{CO}_2 \)

Molar mass of \( \text{CO}_2 \) is \( 12.01 + 2\times16.00 = 44.01 \, \text{g/mol} \).
Moles of \( \text{CO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{14.0 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.318 \, \text{mol} \).

Step2: Relate moles of \( \text{CO}_2 \) to \( \text{Li}_2\text{CO}_3 \)

Assuming the reaction is \( \text{CO}_2 + 2\text{LiOH}
ightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} \), the mole ratio of \( \text{CO}_2 \) to \( \text{Li}_2\text{CO}_3 \) is \( 1:1 \). So moles of \( \text{Li}_2\text{CO}_3 = 0.318 \, \text{mol} \).

Step3: Calculate molecules of \( \text{Li}_2\text{CO}_3 \)

Using Avogadro's number (\( 6.022 \times 10^{23} \, \text{molecules/mol} \)):
Molecules = \( 0.318 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 1.91 \times 10^{23} \, \text{molecules} \).

Answer:

\( 1.91 \times 10^{23} \) molecules (Option: 1.91 x 10^23 molecules)