How many grams of xef6 are required to react with 0.579 l of hydrogen gas at 4.46 atm and 45°c in the reaction shown below?

Hey there!:

Temperature in kelvin : 45 + 273 => 318 K

Number of moles H2:

p * V = n * R * T

4.46 * 0.579 = n * 0.082 * 318

2.58234 = n * 26.076

n = 2.58234 / 26.076

n = 0.09903 moles of H2O

Given the reaction:

XeF6(s) + 3 H2(g) → Xe(g) + 6 HF(g)

Molar  mass XeF6 = 245.3 g/mol

0.09903 mol H2O *  ( 1 mol XeF6 / 3 mol H2 ) * 245.3

= 8.09 g of XeF6

Hope that helps