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:
Mass of CO2 Absorbed by 1.85 g of CaO:
We are given the reaction:
CaO (s)+CO2(g)→CaCO3(s)\text{CaO (s)} + \text{CO}_2 \text{(g)} \to \text{CaCO}_3 \text{(s)}CaO (s)+CO2(g)→CaCO3(s)
From the balanced equation, we know that 1 mole of CaO reacts with 1 mole of CO2 to produce 1 mole of CaCO3. So, the molar ratio between CaO and CO2 is 1:1.
Step 1: Molar Mass of CaO
The molar mass of calcium oxide (CaO) is:
- Ca=40.08 g/mol\text{Ca} = 40.08 \, \text{g/mol}Ca=40.08g/mol
- O=16.00 g/mol\text{O} = 16.00 \, \text{g/mol}O=16.00g/mol
Thus, the molar mass of CaO is:
40.08+16.00=56.08 g/mol40.08 + 16.00 = 56.08 \, \text{g/mol}40.08+16.00=56.08g/mol
Step 2: Moles of CaO in 1.85 g
The number of moles of CaO in 1.85 g is calculated as:moles of CaO=mass of CaOmolar mass of CaO=1.85 g56.08 g/mol≈0.03296 mol\text{moles of CaO} = \frac{\text{mass of CaO}}{\text{molar mass of CaO}} = \frac{1.85 \, \text{g}}{56.08 \, \text{g/mol}} \approx 0.03296 \, \text{mol}moles of CaO=molar mass of CaOmass of CaO=56.08g/mol1.85g≈0.03296mol
Since the reaction occurs in a 1:1 molar ratio, the moles of CO2 absorbed will also be 0.03296 mol0.03296 \, \text{mol}0.03296mol.
Step 3: Mass of CO2 Absorbed
The molar mass of CO2 is:
- C=12.01 g/mol\text{C} = 12.01 \, \text{g/mol}C=12.01g/mol
- O=16.00 g/mol\text{O} = 16.00 \, \text{g/mol}O=16.00g/mol (for 2 oxygen atoms)
So, the molar mass of CO2 is:
12.01+(2×16.00)=44.01 g/mol12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol}12.01+(2×16.00)=44.01g/mol
Now, the mass of CO2 absorbed is:mass of CO2=moles of CO2×molar mass of CO2=0.03296 mol×44.01 g/mol≈1.45 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} \approx 1.45 \, \text{g}mass of CO2=moles of CO2×molar mass of CO2=0.03296mol×44.01g/mol≈1.45g
Thus, 1.85 g of CaO can absorb approximately 1.45 g of CO2.
Volume of CO2 at STP:
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L. Therefore, the volume of CO2 absorbed by 0.03296 mol at STP is:Volume of CO2=moles of CO2×22.4 L/mol=0.03296 mol×22.4 L/mol≈0.74 L\text{Volume of CO2} = \text{moles of CO2} \times 22.4 \, \text{L/mol} = 0.03296 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.74 \, \text{L}Volume of CO2=moles of CO2×22.4L/mol=0.03296mol×22.4L/mol≈0.74L
Thus, the CO2 would occupy approximately 0.74 L at STP.
Summary:
- Mass of CO2 absorbed: 1.45 g
- Volume of CO2 at STP: 0.74 L
