Question

In the following equation, how many grams of CO2 are produced if 2 moles of O2 react completely with methane (CH4)? CH4 + 2O2 -> CO2 + 2H2OHELPFUL INFO: molar mass of CO2=44g/mol, molar mass of O2=32g/mol, molar mass of CH4=16g/mol, and molar mass of water = 18g/mol

Answer 1

Step 1 - "Reading" the equation

The given chemical equation is:

`CH_(4(g))+2O_(2(g))\to CO_(2(g))+2H_2O_((l))`

The bigger numbers, those that come before the formulas of the substances, indicate the quantity in moles of each substance required in this rection.

We can "read" this reaction thus as :

one mole of CH4 reacts with 2 moles of O2 thus producing one mole of CO2 and two moles of H2O

As the exercise is specifically asking about the proportion between O2 and CO2, we can further simplify this statement to:

two moles of O2 produce one mole of CO2

Step 2 - Obtaining the "recipe" for the reaction

In order to obtain a relation in grams, we need to multiply the number of moles of each substance by its respective molar mass. Let's remember that:

two moles of O2 produce one mole of CO2

Therefore, converting to grams (molar masses: 32 g/mol for O2; 44 g/mol for CO2)

`\begin{gathered} O_2\to2\text{ moles }*32\text{ g/mole = 64g} \\ CO2\to1\text{ mole }*44\text{ g/mole = 44g} \end{gathered}`

We have now obtained the "recipe" for this reaction:

64g of O2 produce 44g of CO2

Pretty much as we would with a cake recipe, we can use this to predict how much CO2 is formed.

Since the reaction of two moles of O2 is exactly what we have already calculated, 44g of CO2 would be formed in this reaction.