How many moles of each product?

Question

For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume that there is more than enough of the other reactant.

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)

1. 4.5 mol O2 =mol CO2
2. 4.5 mol O2 =mol H2O
3. 0.0531 molO2 =Mol CO2
4. o.o531 mol O2 =Mol H2O

Answer 1

Hello, Angel,

We are provided a balanced reaction for the combustion of propane and asked to predict the amount of a product formed with a specified amount of one reactant. We are told to that there is a sufficient amount of the second reactant to consume all of the specified reactant.

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)

This equation tells us that if we have 1 mole of C3H8(g) that it will react with 5 moles of O2(g) to produce 3 moles of CO2(g) and 4 moles of H2O(g). [There is also a large amount of heat released from this reaction, but that will come in the thermodynamics chapter].

We can use the molar coefficients as conversion factors. For example, 1 mole of C3H8 will produce 3 moles of CO2. This can be written as:

(3 moles CO2)/(1 mole C3H8)

Another example would be:

(3 moles CO2)/(4 mole H20)

This approach be used to determine the answers:

1. 4.5 mol O2 =mol CO2
2. From the balanced equation, we see that (3 moles CO2/5 moles O2)
3. (3 moles CO2/5 moles O2)*(4.5 mol O2) = (3*4.5/5) moles CO2 [The moles O2 unit cancels out]
4. = 2.7 moles CO2
5. 4.5 mol O2 =mol H2O
6. From the balanced equation, we see that (4 moles H2O)/(5 moles O2)
7. (4 moles H2O)/(5 moles O2)*(4.5 moles O2) = (4*4.5/5 moles H2O)
8. = 3.6 moles H2O
9. 0.0531 mol O2 =Mol CO2
10. Follow the same process. Determine how many moles of CO2 will be produced
11. (3 CO2)/(5 O2)
12. (0.0531 mol O2)*(3 CO2)/(5 O2) = 0.0425 moles CO2
13. 0.0531 mol O2 =Mol H2O
14. (4 H2O)/5 O2)
15. (0.0531 mol O2)*(4 H2O)/5 O2) = 0.0245 moles H2O

I left off the units in problems 3 - 5 to keep it simple. But it is useful to see how they cancel to determine how to write the conversion factors properly.

I hope this helps. It is easier than it looks with a problem that is stated in moles. The same questions when given the amounts in grams requires more effort, but it is the same principle. When one of the reagents is in short supply, the "limiting reagent," the answers need to be adjusted to account for the moles of limiting reagent.