The complete combustion of any hydrocarbon, including methane, produces one water molecule for each two atoms of hydrogen in the hydrocarbon. The formula of methane is CH4; therefore, the complete combustion of one mole of methane produces two moles of H2O.
The combustion reaction of methane is CH4 + 2 O2 = CO2 + 2 H2O. By the law of proportions, there are two molecules of water made for every molecule of methane burned.
Two molecules of water + one molecule of carbon dioxide = three molecules
CH4 + 2O2 = CO2 + 2H2O
2 moles
2
0.750 mol
Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.
The result will be 1 m3 of CO2 and 2 m3 of H2O gas (and 2 m3 of O2 will be consumed).This is determined by the stoichiometry of the balanced reaction:CH4 + 2O2 ---> CO2 + 2H2O
there are two moles produced in potassium nitrate.
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
No moles of oxygen are produced by complete combustion of propane. Oxygen is CONSUMED, not produced. For combustion of 4 moles of propane, it will use 20 moles of oxygen.
1.28mol
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 undergoes complete combustion, the equation for the reaction is CH4 + 2 O2 -> CO2 + 2 H2O. This shows that the number of moles of carbon dioxide formed are the same as the number of moles of methane reacted, so that 14 moles of carbon dioxide will be formed from 14 moles of 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.
0.750 mol
Methane is CH4. Combustion is CH4 + 2O2 ==> CO2 + 2H2O1 mole CH4 produces 2 moles H2Omoles CH4 used = 1.1x10^-3 g x 1 mole/16 g = 6.875x10^-5 molesmoles H2O produced = 6.875x10^-5 moles CH4 x 2 moles H2O/mole CH4 = 1.375x10^-4 molesmass H2O produced = 1.375x10^-4 moles x 18 g/mole = 2.475x10^-3 g = 2.48 mg (3 sig.figs)
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.
The complete combustion of methane proceeds according to the equation: CH4 + 2 O2 = CO2 + 2 H2O. Therefore, each mole of methane produces two moles of water and 4 moles of methane will produce eight moles of water.
Propane is C3H8 and the combustion equation is C3H8 + 5O2 ==> 3CO2 + 4H2OSo the complete combustion of 1 mole of propane requires 5 moles of oxygen.
10,55 moles of water are obtained.
1 mole