Trying to understand how to work this Chemistry problem

Question

A chemistry student was given the following reaction:

C3H8    +     5O2    3CO2    +      4H2O

and was given 50 grams of C3H8 and 72 grams of O2.  The student indicated that this is obviously a limiting reagent problem since two reactant quantities were provided.  Without doing any calculations, the student immediately concluded that the C3H8 must be the limiting reagent because the mass of it is less than the mass of the O2.

Answer 1

Katie

The key point here is the number of moles, not the masses.

C has atomic weight 12.011

H has atomic weight 1.00797

O has atomic weight 15.9994

So C3H8 has atomic weight 3 x 12.011 + 8 x 1.00797 = 44.09676 g

and O2 has atomic weight 2 x 15.9994 = 31.9988 g

So 1 mole of C3H8 (about 6.02 x 10²³ molecules) has mass 44.09676 g

So we have 50 / 44.09676 = 1.13 moles of C3H8

and 72 / 31.9988 = 2.25 moles of O2

For every mole of C3H8 we need 5 moles of O2 to react. So we don't have enough O2 to react with the 1.13 moles of C3H8 ( we need 5 x 1.13 = 5.65 and we only have 2.25).

So what will happen is all the O2 will react but only 2.25 / 5 = 0.45 moles of the C3H8

So actually the O2 is the limiting factor as we do not have enough of it for all the C3H8 to react.

Mike