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If 8.00g of methane is burned with 6.00g of oxygen. How much methane,oxygen,carbon dioxide,and water remain after the reaction is complete in grams?

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This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
CH4 + 2O2 ==> CO2 + 2H2O

Convert 8.00g CH4 to mols. mol = grams/molar mass = about 0.500 g but you should confirm that as well as all of the other numbers. I estimate as I go along.

Convert 6.00 g O2 to mols. 6/32 = about 0.19 mols.

Using the coefficients in the balanced equation, convert mols CH4 to grams CO2.
That is 0.500 mols CH4 x (1 mol CO2/1 mol CH4) = 0.500 x 1/1 = 0.500 mol CO2.

Do the same for mols O2 to mols CO2.
0.19 x (1 mol CO2/2 mols O2) = 0.09 mols CO2.

This gave us two answers for mols CO2 and obviously one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, 0.19 mol CO2 is formed and O2 is the limiting reagent.

Using that value, calculate mols H2O formed. Using that value calculate mols CH4 used. CH4 remaining = initial mols - mols used.

You can convert mols CH4, CO2, and H2O to grams by g = mols x molar mass. There is no oxygen left.
Post your work if you stuck.

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