Ca(OH)? is used to remove CO? during general anesthesia to prevent CO? poisoning: Ca(OH)?(s) + CO?(g) ? CaCO?(s) + H?O(l) a. If 12.0 g of CO? are exhaled and absorbed by 12.0 g Ca(OH)?, how many grams of CO? and Ca(OH)? are left over at the end of the reaction? b. To ensure that all the CO? is captured and removed, an excess amount of Ca(OH)? is generally used. If we want 12.0 g of CO? to only consume 10% of the available Ca(OH)?, what mass of Ca(OH)? should be used?
The Correct Answer and Explanation is:
Let’s break this down step by step.
The reaction is: Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)\text{Ca(OH)}_2 (s) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) + \text{H}_2\text{O} (l)Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)
Part (a): Grams of CO₂ and Ca(OH)₂ left over
To determine the amount of CO₂ and Ca(OH)₂ left over, we first need to balance the equation and calculate how much of each substance will react.
Step 1: Molar masses of the compounds involved
- Molar mass of Ca(OH)₂ = 74.09 g/mol
- Molar mass of CO₂ = 44.01 g/mol
Step 2: Convert the given masses to moles
We are given 12.0 g of each substance, so we can calculate the moles.
For Ca(OH)₂: moles of Ca(OH)2=12.0 g74.09 g/mol=0.162 mol\text{moles of Ca(OH)}_2 = \frac{12.0 \, \text{g}}{74.09 \, \text{g/mol}} = 0.162 \, \text{mol}moles of Ca(OH)2=74.09g/mol12.0g=0.162mol
For CO₂: moles of CO2=12.0 g44.01 g/mol=0.273 mol\text{moles of CO}_2 = \frac{12.0 \, \text{g}}{44.01 \, \text{g/mol}} = 0.273 \, \text{mol}moles of CO2=44.01g/mol12.0g=0.273mol
Step 3: Identify the limiting reagent
From the balanced equation, we see that 1 mole of Ca(OH)₂ reacts with 1 mole of CO₂. So, the limiting reagent will be the substance that is present in the lesser amount of moles.
- We have 0.162 mol of Ca(OH)₂ and 0.273 mol of CO₂.
- Since Ca(OH)₂ has fewer moles, it is the limiting reagent.
Step 4: Calculate the amount of CO₂ that reacts
Since Ca(OH)₂ is the limiting reagent, all of it will react with CO₂. For every 1 mole of Ca(OH)₂, 1 mole of CO₂ will be consumed.
Thus, 0.162 mol of CO₂ will react with 0.162 mol of Ca(OH)₂.
Step 5: Calculate the remaining CO₂
We started with 0.273 mol of CO₂, and 0.162 mol will react, so the remaining CO₂ will be: Remaining CO2=0.273 mol−0.162 mol=0.111 mol\text{Remaining CO}_2 = 0.273 \, \text{mol} – 0.162 \, \text{mol} = 0.111 \, \text{mol}Remaining CO2=0.273mol−0.162mol=0.111mol
Now, convert this back to grams: Remaining CO2=0.111 mol×44.01 g/mol=4.88 g\text{Remaining CO}_2 = 0.111 \, \text{mol} \times 44.01 \, \text{g/mol} = 4.88 \, \text{g}Remaining CO2=0.111mol×44.01g/mol=4.88g
Step 6: Calculate the remaining Ca(OH)₂
Since 0.162 mol of Ca(OH)₂ reacts with 0.162 mol of CO₂, and we started with 0.162 mol of Ca(OH)₂, all of it will be consumed.
Therefore, no Ca(OH)₂ will be left over.
Part (b): Mass of Ca(OH)₂ for 10% reaction
We want to ensure that 12.0 g of CO₂ consumes only 10% of the available Ca(OH)₂.
Step 1: Calculate moles of CO₂
We know from earlier: moles of CO2=12.0 g44.01 g/mol=0.273 mol\text{moles of CO}_2 = \frac{12.0 \, \text{g}}{44.01 \, \text{g/mol}} = 0.273 \, \text{mol}moles of CO2=44.01g/mol12.0g=0.273mol
Step 2: Calculate the moles of Ca(OH)₂ required
The balanced equation tells us that 1 mole of Ca(OH)₂ reacts with 1 mole of CO₂, so 0.273 mol of CO₂ will need 0.273 mol of Ca(OH)₂.
Step 3: Find the total moles of Ca(OH)₂
If 0.273 mol of Ca(OH)₂ is only 10% of the total amount available, then the total moles of Ca(OH)₂ needed will be: Total moles of Ca(OH)2=0.273 mol0.10=2.73 mol\text{Total moles of Ca(OH)}_2 = \frac{0.273 \, \text{mol}}{0.10} = 2.73 \, \text{mol}Total moles of Ca(OH)2=0.100.273mol=2.73mol
Step 4: Convert moles of Ca(OH)₂ to mass
Now, we calculate the mass of Ca(OH)₂ required: Mass of Ca(OH)2=2.73 mol×74.09 g/mol=202.5 g\text{Mass of Ca(OH)}_2 = 2.73 \, \text{mol} \times 74.09 \, \text{g/mol} = 202.5 \, \text{g}Mass of Ca(OH)2=2.73mol×74.09g/mol=202.5g
Final Answers:
- Part (a): 4.88 g of CO₂ will be left over, and no Ca(OH)₂ will remain.
- Part (b): To ensure that only 10% of Ca(OH)₂ is used to capture 12.0 g of CO₂, 202.5 g of Ca(OH)₂ should be used.
