25.5 g of methane burns in the combustion reaction below. how much water is produced in this reaction

CH4 + O2 ===> H2O + CO2 Bare Equation. This is a Combustion Reaction.
CH4 + 2O2>2H2O + CO2 Balanced equation. You start with this.

Step One

Find the moles of Methane.
1 mole of methane has a mass of
C = 12
O2 = 2 * 16 = 32
1 mol of  CO2 = 12 + 32 = 44 grams / mole

Step Two 

Calculate the number of moles of methane.
given mass (g) = 25.5 grams
molar Mass = 44

mol = given mass / molar mass
mol = 25.5 / 44
mol = 0.5795 mols methane.

Step Three

Find the moles of water.
For ever mole of methane, 2 moles of water are produced.

1 mol methane. . . . . . .. .0.5795 mols methane
. . . . .= .  . . . . . . 
2 mol water  . . . . . . . . . . .. .. . .x

Mole of water = 2 * 0.5795
mole of water = 1.159 

Step Four

Find the mass of water.
1 mol = 2H + O = 2 + 16 = 18 grams / mol

desired mass = mols * Molar mass
desired mass = 1.159 * 18
desired mass = 20.86 grams of water.

If you do these questions by dimensional analysis, this problem can be done all in 1 step. It would look like this.

25.5 g [1mol /44grams]*[2mol water / 1 mol meth]*[18 grams / 1 mol water]

There is no reason to think the results are biased or not conducted properly but you should not assume the results are correct just because they were done in a laboratory so the answer would be c. it is likely to be reliable because the results support the conclusion presented.