Question

How many molecules of XeF6 are formed from 12.9 L of F2(at 298 K and 2.60 atm) according to the following reaction?

Assume that there is excess Xe.

Xe(g) + 3 F2(g) → XeF6(g)

Answer 1

Using the ideal gas law , pV = nRT

where p is pressure in atm

V is volume in L = 12.9 L

n is number of moles

R is universal gas constant = 0.0821 L.atm / mol.K

T is temperature in kelvin

so putting all the values , n can be determined for F2

2.60 x 12.9 = n x 0.0821 x 298

n = 1.37 moles

Now 3 moles of F2 produces 1 mole of XeF6 according to the reaction mentioned

So 1.37 moles of F2 will produce 1.37 / 3 = 0.457 moles of XeF6.

1 mole has 6.022 x 10²³ molecules

So 0.457 moles will have 0.457 x 6.022 x 10²³

= 2.75 x 10²³ molecules