Answers:
A. 1333 Â g COâ‚‚; 546.0 g Hâ‚‚O
B. 2200 g COâ‚‚; 370 Â Â g Hâ‚‚O
C. Â 460 L
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place. Â
M_r: Â 56.11 Â Â Â Â Â Â Â Â Â 44.01 Â Â 18.02
     C₄H₈ + 6O₂ ⟶ 4CO₂ + 4H₂O
m/g: Â 425.0
A. Theoretical yield of each product
(1) Calculate the moles of C₄H₈
n = 425.0 ×1/56.11
n = 7.574 mol C₄H₈
(2) Calculate the moles of COâ‚‚ and Hâ‚‚O
In each case, the molar ratio is 4 mol product/1 mol C₄H₈.
COâ‚‚ and Hâ‚‚O:
n = 7.574× 4/1
n = 30.30 mol
(3) Calculate the masses of COâ‚‚ and Hâ‚‚O
Mass of CO₂ = 30.30 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
Mass of COâ‚‚ = 1333 g COâ‚‚
Mass of H₂O = 30.30 mol H₂O × (44.01 g H₂O/1 mol H₂O)
Mass of Hâ‚‚O = 546.0 g Hâ‚‚O
B. Actual yield of each product
Mass of CO₂ = 3333 g × 67/100
Mass of COâ‚‚ = 2200 g COâ‚‚
Mass of H₂O = 546.0 g × 67/100
Mass of Hâ‚‚O = 370 g Hâ‚‚O
C. Volume of gas
At STP, COâ‚‚ is a gas, but Hâ‚‚O is a liquid.
Moles of gas = moles of COâ‚‚
If the actual yield is 67 %
Moles of CO₂ = 30.30 mol × 67/100
Moles of COâ‚‚ = 20 mol
STP is 1 bar and 0 °C.
The molar volume at STP is 22.71 L.
∴ V = 201 mol × 22.71 L/1 mol
  V = 460 L
