Question

100 points help pleaseeee:

Oxygen and hydrogen combine with a lot of heat or a spark, which provides sufficient activation energy, to produce water.

O2(g)+2H2(g)⟶2H2O(l)

Assume 0.290 mol O2
and 0.911 mol H2
are present initially.

After the reaction is complete, how many moles of water are produced?

H2O:

mol
How many moles of hydrogen remain?

H2:

mol
How many moles of oxygen remain?

O2:

mol

What is the limiting reagent?


oxygen


hydrogen

Answer 1

First we need to know which is the limiting agent: wee need to divide the moles of reactants available with its corresponding stoichiometric coefficients. The reactants which ratio is the least is the limiting reagent since less substance can perform the reaction.

O2

0,290 mol / 1 = 0,290

H2

0,991 mol / 2 = 0,456

In this case the limiting agent is oxygen since the ratio si smaller than the hydrogen one.

Since oxygen is the limiting agent, no moles of O2 will remain when the reaction is completed.

Since oxygen is the limiting agent, stoichiometric calculation must be done considering oxygen and not hydrogen. Therefore the amount of water produced will be

`n_(H2O) = 0,290 mol × 2 = 0,580 mol`

And the amount of hydrogen remaining is the subtraction between the hydrogen that has reacted and the total hydrogen available.

Reacted hydrogen:

`n_(H2) = 0,290 mol × 2 = 0,580 mol`

Remaining hydrogen:

`n_(H2) = 0,991 mol - 0,580 mol = 0,411 mol`