Question

ground control reports that 38.5 grams of lioh have been used in the scrubbers since the astronauts left. how many grams of li₂co₃ could be produced by this amount of lioh? show all of your work. you must be able to prove your answer. 23.95 grams 59.48 grams 73.89 grams 118.96 grams

Explanation:

Step1: Determine the molar mass of LiOH

The molar mass of Li (lithium) is approximately 6.94 g/mol, O (oxygen) is 16.00 g/mol, and H (hydrogen) is 1.01 g/mol. For LiOH, the molar mass $M_{LiOH}$ is calculated as:
$$M_{LiOH} = 6.94 + 16.00 + 1.01 = 23.95 \, \text{g/mol}$$

Step2: Calculate moles of LiOH

Given the mass of LiOH is 38.5 grams. Using the formula $n = \frac{m}{M}$, where $n$ is moles, $m$ is mass, and $M$ is molar mass:
$$n_{LiOH} = \frac{38.5 \, \text{g}}{23.95 \, \text{g/mol}} \approx 1.6075 \, \text{mol}$$

Step3: Determine the molar mass of $\boldsymbol{Li_2CO_3}$

For $Li_2CO_3$, the molar mass $M_{Li_2CO_3}$: Li (2 atoms) = $2 \times 6.94 = 13.88$, C = 12.01, O (3 atoms) = $3 \times 16.00 = 48.00$. So,
$$M_{Li_2CO_3} = 13.88 + 12.01 + 48.00 = 73.89 \, \text{g/mol}$$

Step4: Use stoichiometry (assuming a reaction where 2 moles of LiOH produce 1 mole of $Li_2CO_3$)

The balanced reaction (e.g., reaction with $CO_2$: $2LiOH + CO_2
ightarrow Li_2CO_3 + H_2O$) shows 2 moles of LiOH produce 1 mole of $Li_2CO_3$. So moles of $Li_2CO_3$:
$$n_{Li_2CO_3} = \frac{1}{2} \times n_{LiOH} = \frac{1}{2} \times 1.6075 \approx 0.80375 \, \text{mol}$$

Step5: Calculate mass of $\boldsymbol{Li_2CO_3}$

Using $m = n \times M$ for $Li_2CO_3$:
$$m_{Li_2CO_3} = 0.80375 \, \text{mol} \times 73.89 \, \text{g/mol} \approx 59.48 \, \text{g}$$

Answer:

59.48 grams