Answer:

Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆

(b) Moles of CO₂

(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume

Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL