When methane is burned with oxygen, the products are carbon dioxide and water, If you produce 20 grams of water and 10 grams of carbon dioxide from 15 grams of oxygen, how many grams of methane were needed for the reaction?​

4 g

Explanation:

First you need to write a balanced chemical equation. You are given thatmethane is burned, meaning a combustion reaction in which carbon dioxide and water are released.

Unbalanced: CH4 + O2 ---> CO2 + H2O

Balanced: CH4 + 2O2 ---> CO2 + 2H2O

Givens:

X grams CH4 (Molecular mass 16.0 grams)

9 grams H2O (Molecular mass 18.0 grams)

11 grams CO2 (Molecular mass 44.0 grams)

Mole ratio: 1:2:1:2 (CH4:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.

11 g CO2 / (44.0 g) = 0.25 moles CO2

9 g H2O / (18.0 g) = 0.5 moles

n of CH4 = n of CO2 = n of H2O /2 = 0.25 moles

m of CH4 = n* Mw = 0.25 * 16.0 = 4 g

the dependant variable is spilled water

the independent is the heat