Answer:
3,383 Liters H₂(g)
Explanation:
Given Rxn ...
CCl₂F₂(g) + 4H₂(g) => CH₂F₂(g) + 2HCl(g)
?Vol H₂(g) needed to react with 1 Ton CCl₂F₂(g) at 33.5°C (=306.5K) & 225atm
Rxn Ratio CCl₂F₂(g) : H₂(g) => 1 mole CCl₂F₂(g) to 4 moles H₂(g)
1 Ton CCl₂F₂(g) = 2000 lbs CCl₂F₂(g) = 908,000 grams CCl₂F₂(g)
908,000 grams CCl₂F₂(g) = (908,000 grams / 120 grams/mole) = 7,567 moles CCl₂F₂(g)
7,567 moles CCl₂F₂(g) requires 4 x 7,567 moles H₂(g) = 30,267 moles H₂(g)
Using Ideal Gas Law => P·V = n·R·T => V = n·R·T/P
P = 225 atm
n = 30,267 moles H₂(g)
R = 0.08206 L·atm/mol·K
T = 33.5°C = 306.5K
∴Vol H₂(g) needed = (30,267 moles H₂(g))(0.08206L·atm/mol·K)(306.5K)/(225atm) = 3,383 Liters H₂(g)
