How many grams of xef6 are required to react with 0. 579 l of hydrogen gas at 2. 46 atm and 45°c in the reaction shown below?.

At 2.46 atm and 45° C = 318.2 K, we have by the ideal gas law

(2.46 atm) (0.579 L) = n (0.08206 L•atm/(mol•K)) (318.2 K)

where n is the amount (mol) of H₂ present in the sample. Solve for n :

n = (2.46 atm) (0.579 L) / ((0.08206 L•atm/(mol•K)) (318.2 K))

n ≈ 0.0545 mol

Then depending on the "reaction shown below", you should have some indication of the ratio of moles of XeF₆ to moles of H₂. Find the number of moles of XeF₆ and convert that accordingly to a mass (the molar mass is about 245.283 g/mol).

probably a 9

explanation: