Methane burns in the presence of oxygen to form carbon dioxide and water. 2nd attempt
A sealed chamber contains 7.50 g CH4 and 13.5g 02.
The mixture is ignited producing CO2 and H20. How many grams of CO2 are produced?

= 9.28 g CO₂

Explanation:

First write a balanced equation:

CH₄ + 2O₂ -> 2H₂O + CO₂

Convert the information to moles

7.50g CH₄ = 0.46875 mol CH₄

13.5g O₂ = 0.421875 mol O₂

Theoretical molar ratio CH₄:O₂ -> 1:2

Actual ratio is  0.46875 : 0.421875 ≈ 1:1

If all CH₄ is used up, there would need to be more O₂

So O₂ is the limiting reactant and we use this in our equation

Use molar ratio to find moles of CO₂

0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂

Then convert to grams

0.2109375 mol CO₂ = 9.28114 g CO₂

round to 3 sig figs

= 9.28 g CO₂

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