One of the few xenon compounds that form is cesium xenon heptafluoride (csxef7). how many moles of csxef7 can be produced from the reaction of 11.0 mol cesium fluoride with 11.5 mol xenon hexafluoride?
csf(s) + xef6(s) csxef7(s)

11 moles

Explanation:

As can be seen from the balanced chemical reaction,

1 mole of CsF reacts with 1 mole of

Thus 11 moles of CsF react with 11 mole of

Thus CsF is the limiting reagent as it limits the formation of products and is the excess reagent as (11.5-11)=0.5 moles are in excess.

As  1 mole of CsF produces= 1 mole of

11 moles of CsF will produce= of

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.