To solve this problem, we need to determine the volume of carbon dioxide (CO₂) that can be scrubbed by 2 kilograms of lithium hydroxide (LiOH) at standard conditions. We will do this in a step-by-step manner as follows:
1. Determine the molar mass of LiOH:
- The molar mass of lithium hydroxide (LiOH) is 23.94 grams per mole (g/mol).
2. Convert the mass of LiOH to moles:
- Given mass of LiOH is 2 kilograms, which is equivalent to 2000 grams.
- Moles of LiOH needed = [tex]\( \frac{\text{mass of LiOH}}{\text{molar mass of LiOH}} \)[/tex]
- Moles of LiOH needed = [tex]\( \frac{2000 \text{ grams}}{23.94 \text{ g/mol}} \)[/tex]
- This calculation results in approximately 83.5422 moles of LiOH.
3. Use the stoichiometry of the reaction:
- According to the balanced chemical equation, 2 moles of LiOH react with 1 mole of CO₂.
- Moles of CO₂ scrubbed = [tex]\( \frac{\text{moles of LiOH}}{2} \)[/tex]
- Moles of CO₂ scrubbed = [tex]\( \frac{83.5422}{2} \)[/tex]
- This gives us approximately 41.7711 moles of CO₂.
4. Convert moles of CO₂ to volume at standard conditions:
- At standard conditions (STP), 1 mole of any gas occupies 22.4 liters.
- Volume of CO₂ scrubbed = moles of CO₂ scrubbed × 22.4 liters/mole
- Volume of CO₂ scrubbed = 41.7711 × 22.4
- This results in approximately 935.6725 liters of CO₂.
Thus, 2 kilograms of lithium hydroxide can scrub or remove approximately 935.6725 liters of carbon dioxide at standard conditions.
