Question

Why is this statement false: "If 10 grams of propane burn, you need 50 grams of oxygen."?

Answer 1

Well, let us interrogate the stoichiometric equation.........

Explanation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)`

And thus by stoichiometry, approx. `160*g` of dioxygen are required to combust `44*g` of propane completely. From where am I getting these masses?

You have specified `10*g` of propane................

i.e. `"Moles of propane"` `=` `(10*g)/(44.1*g*mol^-1)=0.227*mol`

And thus we require `5xx0.227*molxx32.0*g*mol^-1=36.3*g` dioxygen gas........

Answer 2

To find the mass of a reactant/product from that of another reactant/product, it is usually necessary to go via a mole ratio .

Explanation:

Firstly, observe the equation for the complete combustion of propane:

`"CH"_3"CH"_2"CH"_3 + 5"O"_2 -> 3"CO"_2 + 4"H"_2"O"`

From this, it is clear that there is a 1:5 ratio between propane and oxygen for complete combustion; however, this relationship is not one of mass, but of amount - in other words, a mole ratio.

Consider the moles of propane being combusted:

`"mol(CH"_3"CH"_2"CH"_3")"=("mass(CH"_3"CH"_2"CH"_3")")/("M"_r"(CH"_3"CH"_2"CH"_3")")=10/44.0~~0.22727" moles"`

Then apply the mole ratio we saw above to find the number of moles of oxygen gas required.

`"mol(O"_2")"=5*"mol(CH"_3"CH"_2"CH"_3")"=5*0.22727~~1.13636`

From this, the mass of oxygen gas required can be calculated:

`"mass(O"_2")"="mol(O"_2")"*"M"_r"(O"_2")"=1.13636*32.0=36.dot3dot6`

So the mass of oxygen gas required would be `36"g"` to 2 significant figures.

It is important to recognise the difference between mass and amount. For example, two metals involved in a single-displacement reaction may have the same charge (and so react in a 1:1 ratio), but they will undoubtedly have different relative atomic masses. This means that, although the amount of solid metal will remain the same, there will be an observed mass difference before and after such a process.

In summary, to find the mass of a reactant/product from that of another reactant/product, it is usually necessary to go via a mole ratio.