I need help with this chemistry question, please explain your answers if it isn't too much trouble.

Question

The methane CH4 of 29.0 L undergoes complete combustion at 0.961 atm and 140 C.

a. Write down the balanced equation of the chemical reaction producing carbon dioxide

and water vapor.

b. How many liters of CO 2 are produced?

c. How many liter of H 2 O are produced?

If 64 grams of O 2 was used to burn 20 grams of CH 4

d. How many grams of CO 2 were produced?

e. CH 4 were completely combusted?

Answer 1

(a). CH₄ + 2O₂ ==> CO₂ + 2H₂O ... balanced equation for combustion of methane

(b). 29.0 L CH₄ x 1 L CO₂ / L CH₄ = 29.0 L CO₂ produced

(c). moles O₂ used = 64 g O₂ x 1 mol O₂/32 g = 2 mol O₂

moles CH₄ used = 20 g CH₄ x 1 mol CH₄ / 16 g = 1.25 mol CH₄

Since they react in a 1:1 mol ratio according to the balanced equation, CH₄ is limiting and determines the amount of product that can be formed. Using this information, and the stoichiometry of the balanced equation, we have...

1.25 mol CH₄ x 2 mol H₂O/mol CH₄ = 2.50 mol H₂O produced. Converting to volume, we have...

V = nRT/P = (2.50 mol)(0.0821 Latm/Kmol)(413K)/0.961 atm = 88.2 L H₂O

(d). 1.25 mol CH₄ x 1 mol CO₂ / mol CH₄ x 44 g CO₂/mol = 55 g CO₂ produced

(e). I don't understand what they are asking. It's a statement, not a question.