Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
5 mol O2 * 1 mol CH4 / 2 mol O2 = 2.5 mol CH4 needed for reaction
Therefore CH4 is the excess reactant and O2 is the limiting reactant
No. Oxygen is a reactant in combustion, not a product.
Oxygen + organic molecule -------> carbon dioxide + water
Glucose
Combustion products of methane would be water and CO2. Methane is CH4. http://en.wikipedia.org/wiki/Methane
Complete combustion of a hydrocarbon yields carbon dioxide & water; incomplete combustion yields carbon monoxide & water. By having excess oxygen you have enough oxygen to ensure complete combustion. For example the combustion of methane (CH4):complete combustion: CH4 + 2O2 --> CO2 + 2H2Oincomplete combustion: CH4 + 1.5O2 --> CO + 2H2OAs you can see you need a 1/2 mole less of oxygen for the incomplete combustion of methane. So as long as you have twice the amount (in terms of moles) of oxygen as methane you will ensure complete combustion. So anything in excess of that will also ensure complete combustion.
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.
Every combustion reaction we deal with produces gas with oxygen in the product, so O2 (oxygen gas) must be a reactant. For example, methane reacts with Oxygen in this way: CH4(l) + O2(g) -> C02(g)+2H2(g) Note O2 in gaseous form as a reactant.
No. Molecular oxygen is a reactant in a combustion reaction.
Oxygen
Oxygen
Combustion. Any reaction that has O2 as a reactant is combustion.
No. Oxygen is a reactant in combustion, not a product.
The limiting reactant is oxygen.
Oxygen.
The important reactant is oxygen.
Oxygen
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?