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)

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error