Question

REACTION; C5H12 + 8O2 5CO2 + 6H2OWhen 15.5 grams of C5H12 are consumed in this reaction what volume of CO2 can be produced in liters?

Answer 1

Step 1 - Finding the stoichiometry of the reaction

The given reaction is:

`C_5H_(12)+8O_2\to5CO_2+6H_2O`

To find the stoichiometry of the reaction in moles, we just have to "read" the bigger numbers that come before the formulas of the substances:

1 mole of C5H12 reacts with 8 moles of O2 thus producing 5 moles of CO2 and 6 moles of H2O

Since the exercise is specifically asking us about the relation between C5H12 and CO2, we can simplify this statement to:

1 mole of C5H12 produces 5 moles of CO2

Step 2 - Converting this relation to a relation in grams

To convert moles to grams, we just need to multiply the number of moles by the molar mass of the substance.

Since the exercise is giving us the mass of C5H12, it would be interesting to convert its number of moles to mass (its molar mass is 72 g/mol):

`C_5H_(12)\to1*72=72g`

We can now rewrite the statement in step 1 as:

72g of C5H12 produce 5 moles of CO2

Step 3 - Finding how many moles of CO2 will be formed

To find the moles of CO2 that will be formed, we can set the following proportion:

`\begin{gathered} 72g\text{ of C5H12 produce ---- 5 moles of CO2} \\ 15.5g\text{ of C5H12 would produce --x } \\ \\ x=(15.5*5)/(72)=1.07\text{ moles of CO2} \end{gathered}`

Step 4 - Converting moles to volume

The molar volume of a gas in STP conditions (standard pressure and volume) is 22.4 L/mol, which means that each mole of gas occupies 1 L. Since we have formed 1.07 moles of CO2, we have:

`V=1.07*22.4=23.96L`

The volume of CO2 that would be formed is thus 23.96L.