Question

In the following combustion reaction you use 43.5 grams of C3H8. How many moles of H2O was produced?C3H8 + 5O2 → 3CO2 + 4H2O

Answer 1

Step 1 - Finding the stoichiometry of the reaction

The given reaction is:

`C_3H_8+5O_2\to3CO_2+4H_2O`

The stoichiometry of the reaction can be found by reading the bigger numbers, those that come before the formulas of the substances:

1 mole of C3H8 react with 5 moles of O2 thus producing 3 moles of CO2 and 4 moles of H2O

Since the exercise is specifically asking us about the proportion between C3H8 and water, we can simplify the statement above to:

1 mole of C3H8 produces 4 moles of H2O

Step 2 - Converting the number of moles of C3H8 to grams

Note that the exercise is asking us about how much water would be produced if we used 43.5 grams of C3H8. For sake of simplicity, let's convert the number of moles of C3H8 to mass (grams).

Converting moles to grams is easy. We just have to multiply the number of moles by the molar mass of the substance (44 g/mol for C3H8):

`C_3H_8\to1*44=44\text{ g}`

We can now rewrite the statement in step 1 as:

44 g of C3H8 produce 4 moles of H2O

Step 3 - Finding how many moles of water would be produced

Now we just have to set a proportion:

`\begin{gathered} 44\text{ g of C3H8 produce --- 4 moles of H2O} \\ 43.5\text{ g of C3H8 would produce -- x} \\ \\ x=(4*43.5)/(44)=3.95\text{ moles of H2O} \end{gathered}`

Therefore, 3.95 moles of H2O would be produced in this reaction.