Question

The combustion reaction of C3H8O is represented by this balanced chemical equation: 2C3H8O + 9O2= 6CO2 + 8H2O. when 6.01×10^22 molecules of O2 and 2.78×12^23 molecules of C3H8O are combined, how many grams of which reagent was used?

Answer 1

0.132g of C3H8O
0.319g of O2

Explanation:

From the balanced chemical equation, the first thing we realize should realize is that for every `2 C3H8O` molecules, we need `9 O2` molecules.
What this means is that in the chemical reaction, the molecules react in this ratio, however, there is a possibility that we could have more `C3H8O` than we need. For example, if you had 5 `C3H8O` molecules and 18`O2` then you have 1 extra `C3H8O` molecule because 4 of them would have reacted with the 18 `C3H8O` molecules but the extra 1 molecule will NOT have reacted.

From the data given to us,
Situation 1
if All `2.78*12^23` molecules of `C3H8O` had reacted, then by the ratio of 2:9, we would expect `8.28 *10^25` molecules of `O2` to have reacted with it.

Situation 2
if ALL `6.01*10^21` molecules of `O2` had reacted, then by the ratio of 2:9, we would expect `1.336 *20^21` molecules of `C3H8O` to have reacted with it.

As you can see, situation 1 is impossible since the number of `O2` molecules needed to react with `C3H8O` is significantly more than what was given to us.

From situation 2, we can see that we have more than enough `C3H8O` needed to react with `O2`. the number of UN-reacted molecules would therefore be `(2.78*12^23)-(1.336 *20^21)`.

Since situation 2 is the only possible situation, you can now calculate the number of grams using their moles.

mass of oxygen = `(6.01*10^21)/(6.02*10^23) *(32)=0.319g`

mass of `C3H8O` = `(1.336 *20^21)/(6.02*10^23) *(60)=0.132g`

32 and 60 are their respective relative atomic masses.