Question

In another 8.20 L flask was filled with 3.60 moles of H2, 1.44 mol of O2 at a temperature of 127°C.The mixture in the flask is ignited by a spark, and the reaction represented below occurs until one of the reactants is entirely consumed.
2H2(g) + O2(g) → 2H2O(g)
Give the mole fraction of all the species present in the flask at the end of the reaction.

Answer 1

First we need to find the limiting reagent, since one of the reactants is entirely consumed - do this by dividing the moles of each reactant by the stoichiometric coefficient in the equation:

For H2: 3.6/2 = 1.8
For O2: 1.44/1 = 1.44

O2 has the lower value, so it will be the limiting reagent. Let's first find out how much H2 will remain. From the reaction, we can see that for however much O2 reacts, twice as much H2 reacts. If all our 1.44 mol of O2 is used up, that means twice as much H2, or 2.88 mol, will also be used up. This leaves us with 3.6 - 2.88 = 0.72 mol of H2 when the reaction is finished!

Now to figure out how much H2O is produced. From the reaction, we can see that for however much O2 reacts, twice as much H2O is produced. If all our 1.44 mol of O2 is used up, that means twice as much H2O, or 2.88 mol, will be produced.

Finally, we need to find mole fractions, so we'll first need to find total moles. At the end of the reaction, we have 0.72 mol of H2 + 2.88 mol of H2O = 3.6 mol of stuff. This means:

Mole fraction of H2: 0.72/3.6 = 0.2
Mole fraction of H2O: 2.88/3.6 = 0.8

If you have any questions please ask!

Answer 2

You are amazing

Answer 3

Haha glad I could help!