Question

At a fixed temperature, equal moles of N₂ (g) and F₂ (g) are mixed in a constant pressure container (the volume of the container changes in order to keep the pressure at a constant value). The N₂ (g) and F₂ (g) are allowed to react, producing NF₃ (g):

N₂ (g) + 3F₂ (g) ? 2 NF₃ (g)

If the initial volume of the container, before any reaction takes place, is 4.10 L, determine the volume of the container after the N₂ (g) and F₂ (g) have reacted to completion.

The final volume will be  L

Answer 1

we can use this equation ,

PV=nRT.because there is change in volume,we have two equations

(P1V1)/(n1t1)=R=(P2V2)/(n2T2) P is pressure, V is volume, n is number of moles, t is temperature in Kelvin, and R is a universal constant.

since pressure and temperature constant , the equation becomes

(V1)/(n1)=(V2/n2)

Since you have equal moles of each, assume that you have 1 mol of both H2 and O2.
The total number of moles then is 2 moles each. Set this as your n1 value.

to find n2,now we should find limiting reagent

the one with lesser number of moles is limiting reagent

instead of 3 moles we have only 1 mole of F2 so it is limiting reagent

3 moles of F2 gives 2 moles of NF_{3 .} for 1 mole of F2 you will get =1molx2mol/3mol=2/3 mol of NF3 will form.

excess of rectant is N2 . 1mole of F2 will react with only 1/3 of N2 the remaining 2/3 of N2 will be in the container with the product

n2=2/3+2/3=1.3333

Now we can solve for v2
(V1/n1)=(V2/n2)

(V1n2)/(n1) = V2

4.10Lx1.3333mol/2mol=2.73L

final volume =2.73L