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

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

the answer is definitely c.