What mass of carbon dioxide is produced from the complete combustion of 6.80103g of methane?

Answer 1

The balanced equation for the complete combustion of methane is CH4 + 2 O2 -> CO2 + 2 H2O. This equation shows that each mole of methane produces one mole of carbon dioxide. The mass of carbon dioxide produced by complete combustion of any mass of methane will therefore equal the mass of methane reacted multiplied by the ratio of the gram molecular masses of carbon dioxide and methane. The answer to the question therefore is about (6.80103)(44.0098)/(16.04276) or 18.6953 grams of carbon dioxide, to the justified number of significant digits.

Answer 2

Combustion of methane (CH4) is : CH4 + 2O2 ---> CO2 + 2H2O

moles CH4 = 5.20103 g x 1 mol/16g = 0.325 moles (you can add sig figs by using exact molar mass)

moles CO2 produced = 0.325 since mole ratio of CO2:CH4 is 1:1

Answer 3

Balanced equation: 4CH3 + 7O2 ---> 4CO2 + 6H2O
moles of CH3 present = 6.20x10^3 g x 1 mol/15 g = 413 moles
413 moles CH3 = 413 moles of CO2 since mole ratio in balanced equation is 1:1
mass of CO2 produced = 413 moles x 44 g/mole = 18,172 g = 18.2 kg

Answer 4

Take the balanced equation CH4+2O2---->CO2+2H2O.16g produces 44g of CO2.So 6.70103g gives 18.427g of CO2.

Answer 5

[3.90X10^-3g CH4 (44.031g CO2)] / 16.042g CH4= 0.0107g CO2

Answer 6

From 6,80 10ex.3 g CH4 37,4 10ex.3 g CO2 are formed.

Answer 7

If 6.20103 is 6 200 grams 17 050 grams of carbon dioxide are produced.

Answer 8

If the mass of methane is in g the mass of carbon dioxide is 21 676 g.