Calcium oxide can be used to “scrub” carbon dioxide from air. CaO (s) + CO2 (g) –> CaCO3 (s) What mass of CO2 could be absorbed by 1.85 g of ? Mass = g CO2 What volume would this CO2 occupy at STP? Volume = L CO2
The Correct Answer and Explanation is:
Step 1: Write the balanced chemical equation
The reaction between calcium oxide (CaO) and carbon dioxide (CO2) is given by:
CaO (s)+CO2 (g)→CaCO3 (s)\text{CaO (s)} + \text{CO2 (g)} \rightarrow \text{CaCO3 (s)}CaO (s)+CO2 (g)→CaCO3 (s)
This shows that 1 mole of calcium oxide reacts with 1 mole of carbon dioxide to form 1 mole of calcium carbonate.
Step 2: Determine moles of CaO
We need to find the moles of CaO in 1.85 g.
- Molar mass of CaO = 40.08 g/mol (for Ca) + 16.00 g/mol (for O) = 56.08 g/mol.
Now calculate the moles of CaO in 1.85 g: moles of CaO=massmolar mass=1.85 g56.08 g/mol=0.03296 mol\text{moles of CaO} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.85 \, \text{g}}{56.08 \, \text{g/mol}} = 0.03296 \, \text{mol}moles of CaO=molar massmass=56.08g/mol1.85g=0.03296mol
Step 3: Moles of CO2 absorbed
From the balanced equation, we know that 1 mole of CaO reacts with 1 mole of CO2. Therefore, the moles of CO2 absorbed by 1.85 g of CaO will be equal to the moles of CaO: moles of CO2=0.03296 mol\text{moles of CO2} = 0.03296 \, \text{mol}moles of CO2=0.03296mol
Step 4: Calculate mass of CO2 absorbed
Now we can calculate the mass of CO2 absorbed. The molar mass of CO2 is: Molar mass of CO2=12.01 g/mol (C)+2×16.00 g/mol (O)=44.01 g/mol\text{Molar mass of CO2} = 12.01 \, \text{g/mol (C)} + 2 \times 16.00 \, \text{g/mol (O)} = 44.01 \, \text{g/mol}Molar mass of CO2=12.01g/mol (C)+2×16.00g/mol (O)=44.01g/mol
The mass of CO2 absorbed is: mass of CO2=moles of CO2×molar mass of CO2=0.03296 mol×44.01 g/mol=1.453 g\text{mass of CO2} = \text{moles of CO2} \times \text{molar mass of CO2} = 0.03296 \, \text{mol} \times 44.01 \, \text{g/mol} = 1.453 \, \text{g}mass of CO2=moles of CO2×molar mass of CO2=0.03296mol×44.01g/mol=1.453g
So, the mass of CO2 absorbed by 1.85 g of CaO is 1.45 g (rounded to 3 significant figures).
Step 5: Calculate the volume of CO2 at STP
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 L. Therefore, we can calculate the volume of CO2 absorbed as: Volume of CO2=moles of CO2×22.4 L/mol\text{Volume of CO2} = \text{moles of CO2} \times 22.4 \, \text{L/mol}Volume of CO2=moles of CO2×22.4L/mol Volume of CO2=0.03296 mol×22.4 L/mol=0.739 L\text{Volume of CO2} = 0.03296 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.739 \, \text{L}Volume of CO2=0.03296mol×22.4L/mol=0.739L
Thus, the volume of CO2 absorbed by 1.85 g of CaO at STP is 0.739 L.
Final Answer:
- Mass of CO2 absorbed: 1.45 g
- Volume of CO2 at STP: 0.739 L
Explanation:
The process begins with determining the moles of CaO, which is then used to find the moles of CO2 based on the 1:1 molar ratio. The molar mass of CO2 is then used to calculate the mass of CO2 absorbed, and finally, the volume of CO2 is calculated using the molar volume at STP. This provides both the mass and the volume of CO2 that can be absorbed by the given amount of calcium oxide.
