CCl2F2 (g) + 4H2 (g) → CH2F2 (g) + 2HCl (g)

What volume of hydrogen at 225atm and 33,5 °C would be required to react with 1 ton of CCl2F2?

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)