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)