Help with a fairly simple Chemistry question!?

Question

2H2 + 02 ---------> 2H2O

Identify the limiting reagent:

50 molecules of H2 and 25 molecules of O2

(My biggest problem was setting the molecules to grams. I can only figure out how to convert moles to grams, not in between.)

Thanks!

Answer 1

It is by definition that a limitng reagent is the reactant that, in a reaction, will be depleted first;  and the lesser number of moles. From the balanced equation, on the reactant side, we see that we require 1 mole of O2 to react with 2 mole of H2. So, O2 is the limiting reagent which is the lesser number of moles.

For this question, you do not require to do any conversion.

Answer 2

Actually, O2 is not the limiting reagent.

The limiting reagent in a reaction is the reagent that will be all used if the reaction goes to completion. It limits the reaction because once all of one reagent is gone then no more reaction can occur.

For the example that you have given above there is no limiting reagent. You have been given 50 molecules of H2 and 25 molecules of O2. The equation tells you that 2 moles of H2 react exactly with 1 moles of O2, which means that , using 50H2 and 25O2, all of both reactants will be used in the reaction. Also, the end product will be 50 molecules of H2O

EXAMPLE 1:

Lets make some numbers up:

O2 = 0.6 moles

H2 = 0.1 moles

remembering that 2 moles of H2 is required to react with one mole of O2 then:

0.6 moles of O2 would need (2 x 0.6) moles of H2 to fully react

= 1.2 moles of H2

Obviously, since we have only 0.1 moles of H2 we have no where near enough H2 to fully react with all the O2. So in this case the limiting reagent is H2.

The 0.1 moles of H2 will react with 0.05 moles of O2

There will be 0.6 - 0.05 moles of O2 left unreacted (0.55 moles)

Since 2 H2 reacts to give 2 H2O, which is a 1:1 ratio, then the result of the reaction will be 0.1 moles of H2O

EXAMPLE 2:

When you are given the amounts of reactants in mass then you must first work out how many moles of each reactant you have, then compare the ratio they react in to work out your answer.

Lets make up some masses

10.0 g of H2

50 g of O2

Now, find the number of moles represented by the mass of each reagent:

The equation is

moles = mass / molecular weight

Moles of H2 = mass / molecular weight

molecular weigth H2 = (2 x 1.008) = 2.016 g/mol

therefore moles H2 = 10.0g / 2.016 g/mol

= 4.96 moles of H2

Moles of O2 = mass / molecular weight

molecular weight O2 = (2 x 16.0) = 32.0 g/mol

therefore moles O2 = 50.0 g/ 32.0 g/mol

= 1.56 moles O2

Now we compare the moles of each.

For 4.96 moles of H2 to fully react we need (1/2 x 4.96) moles of O2

= 2.48 moles.

But we worked out above that we only have 1.56 moles of O2, which is not enough. So in this case O2 is the limiting reagent.

The 1.56 moles of O2 will react with (2 x 1.56) moles of H2

= 3.12 moles of H2 required to react with 1.56 moles of O2

Unreacted H2 = 4.96 - 3.12 =  1.84 moles of H2 left over

To work out the mass of unreacted H2 use

mass unreacted H2 = moles x molecular weigth

= 1.84 moles x 2.016 g/mol

= 3.71 g of H2 is unreacted (in excess)

The amount of water produced from 1.56 moles of O2

1 mole of O2 ----> 2 moles of H2O

therefore 1.56 moles O2 -------> 3.12 moles H2O

Mass of water produced:

molecular mass H2O = (2 x 1.008) + 16.0 = 18.016 g/mol

moles = mass / molecular weight

Therefore mass = moles x molecular weight

= 3.12 mol x 18.016 g/mol

= 56.21 g of water would be produced