When CH4 burns, it reacts with oxygen gas to produce C02 andH2O. The unbalanced equation for this reaction is 1CH4(g)+2O2(g)→1CO2(g)+2H2O(g)?

Mass of CO2 produced:

1.00 x10-3 g methane.. Because mass = mr x moles

mr of methane = 16

1x10-3/16 = 6.25x10-5 moles

6.25x10-5 moles of CO2 are produced

mr of C02 = 44

mr x moles = mass

therefore 6.25x10-5 x 44 = 2.75x10-3 g of C02 are produced

Mass of Water produced:

6.25x10-5 moles of CH4 are burned, therefore 1.25x10-4 moles of H20 are produced (there are two H20 molecules formed therefore twice the number of moles)

Mr H20 = 18

1.25x10-4 x 18 = 2.25 x 10-3 g Of H20 produced

Mass of Oxygen needed:

Once again for Oxygen, 1.25x10-4 moles are needed

mr of 02 = 32

32 x 1.25x10-4 = 4x10-3 g

Hope this helps :) If you remember that mass = moles x mr, you can rearrange this equation to work out almost every question like this

CH4 + 2 O2 →CO2 + 2 H2O

moles CH4: 0.001 g/16 g/mole= 6.25E-5

from bal rxn., 1 mole of CO2 is produced per mole CH4

6.25E-5 moles CO2 x 44 g/mole CO2= 2.75E-3 g CO2

from bal. rxn. 2 moles of H2O are formed per mole CH4, 2:1

6.25E-5 moles CH4 x 2 x 18 g/mole H2O= 2.25E-3 g H2O

from bal. rxn. 2 moles of O2 react with 1 mole CH4, 2:1

6.25E-5 moles CH4 x 2 x 32 g/mole O2= 4E-3 g O2