Answer:
1. 20.5 L of CO₂ are released.
2. 44.8L of O₂ are needed
Explanation:
C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂ (g) + 6H₂O(l)
We assume oxygen as reagent in excess.
Let's convert the mass of glucose to moles → 24 g . 1mol/180 g = 0.133 moles
Ratio is 1:6 so (0.133 . 6) = 0.8 moles of CO₂ are released
We apply the Ideal Gases Law to determine the volume:
P . V = n . R .T → V = (n. R .T)/P
V = (0.8 mol . 0.082 . 310K) /0.990 atm = 20.5L
For 2nd question: First of all, we determine the moles of glucose
55 g . 1mol/180 g = 0.305 moles
We apply a rule of three: 1 mol of glucose needs 6 moles of O₂ to react
0.305 moles may need (0.305 .6)/1 = 1.83 moles
We use the Ideal Gases Law again:
V = (1.83 mol . 0.082 . 298K) / 1 atm = 44.8L
