CH4 + 2 O2 --> CO2 + 2 H2O
x / 64 = 1 / 16
x = 4 g oxygen
Assuming oxygen is 20% oxygen...
4 = .2 x
x = 20 g air that is one possible answer.
A better answer--- air is 20% oxygen by volume and 80% nitrogen by volume.
1g methane / 16 = 0.0625 mole
0.0625 / 1mole meth = x / 2mole oxy
x = 0.125 mole oxygen =20% of air
--- 0.625 mole 'air'
0.2 x 32 + 0.8 x 28 = 28.8 g/mole of air
0.625 x 28.8 = 18 grams of air
Since air is 20% oxygen by volume, the 2nd answer is best.
Oxygen is NOT a PRODUCT (it is not produced) from the complete combustion of methane, it is a REACTANT (it is used in the reaction). The answer is therefore a mass of zero.
According to Lavoisier: Mass reactants [total in] = mass products [total out] x g Methane + 32 g oxygen = 22 g carbon dioxide + 18 water x + 32 = 22 + 18 = 40 x = 40 - 32 = 8 g methane
The mass of CO2 is 1 105 g.
Its conserved during the combustion of anything - the mass of the products is always equal to the mass of the materials that react.
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.
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
The answer is 24,15 g.
Oxygen is NOT a PRODUCT (it is not produced) from the complete combustion of methane, it is a REACTANT (it is used in the reaction). The answer is therefore a mass of zero.
Burning 2 700 g of methane produce 70406 g of carbon dioxide.
The mass of reactants is equal to the mass of products.
According to Lavoisier: Mass reactants [total in] = mass products [total out] x g Methane + 32 g oxygen = 22 g carbon dioxide + 18 water x + 32 = 22 + 18 = 40 x = 40 - 32 = 8 g methane
The mass of CO2 is 1 105 g.
16.9
Its conserved during the combustion of anything - the mass of the products is always equal to the mass of the materials that react.
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.
First a balanced chemical equation is needed.CH4 + 2O2 -> CO2 + 2H2OThere is a 1:1 ratio of moles between methane and carbon dioxide so the amount of moles of methane used is the exact number of moles of carbon dioxide yielded.To determine the number of moles of methane we take the amount used and divide by methane's mass which is about 16.04 g/mol.100g/ 16.04g/mol=6.234moles of methane.6.234 moles of methane are used and 6.234 moles of carbon dioxide are produced.
When methane burns, the carbon dioxide and water formed, equal the mass of the methane plus the mass of the oxygen.