0
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.
What else can I help you with?
How many molecules of oxygen are needed for the incomplete combustion of one molecule of methane?
For the incomplete combustion of one molecule of methane (CH4) to form carbon monoxide (CO) rather than carbon dioxide (CO2), one molecule of oxygen (O2) is needed. This results in the equation: CH4 + O2 -> CO + H2O.
Why is mass conserved during the combution of methane?
Mass is conserved during the combustion of methane due to the principle of conservation of mass, which states that matter cannot be created or destroyed, only transformed. In the case of methane combustion, the reactants (methane and oxygen) are converted into products (carbon dioxide and water) through a chemical reaction. The total mass of the reactants is equal to the total mass of the products, demonstrating the conservation of mass.
What is the amount of water produced by the combustion of 32grams of methane?
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
How many moles of carbon dioxide are produced from the complete combustion of 100.0 grams of methane?
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
What mass of water is produced from the complete combustion of 2.50x10-3 of methane?
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)
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
What mass of carbon dioxide is produced from the complete combustion of 8.80103g of methane?
The answer is 24,15 g.
What mass of carbon dioxide is produced from the complete combustion of 2.70103 g of methane?
Burning 2 700 g of methane produce 70406 g of carbon dioxide.
Why the combined mass of CO2 and H2O produced in the combustion analysis of methane greater than mass of the methane sample being analysed?
The mass of reactants is equal to the mass of products.
How many molecules of oxygen are needed for the incomplete combustion of one molecule of methane?
For the incomplete combustion of one molecule of methane (CH4) to form carbon monoxide (CO) rather than carbon dioxide (CO2), one molecule of oxygen (O2) is needed. This results in the equation: CH4 + O2 -> CO + H2O.
Why is mass conserved during the combution of methane?
Mass is conserved during the combustion of methane due to the principle of conservation of mass, which states that matter cannot be created or destroyed, only transformed. In the case of methane combustion, the reactants (methane and oxygen) are converted into products (carbon dioxide and water) through a chemical reaction. The total mass of the reactants is equal to the total mass of the products, demonstrating the conservation of mass.
Is mass conserved in methane?
Yes, mass is conserved in the combustion of methane. According to the law of conservation of mass, the total mass of reactants (methane and oxygen) equals the total mass of products (carbon dioxide and water) during the reaction. Although the form of the substances changes, no mass is lost or gained in the process.
What is the amount of water produced by the combustion of 32grams of methane?
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
How many moles of carbon dioxide are produced from the complete combustion of 100.0 grams of methane?
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
When methane burns carbon dioxide and water are formed The mass of carbon dioxide plus the mass of water equal the mass of the methane plus the mass of OXYGEN?
When methane burns, the carbon dioxide and water formed, equal the mass of the methane plus the mass of the oxygen.
What mass of water is produced from the complete combustion of 2.50x10-3 of methane?
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)
What mass of O2 is required for the complete combustion of 7.91 g of C3H8 to CO2 and H2O?
28.76 g of O2 Write the complete combustion reaction and balance C3H8 + 5O2 yields 3CO2 + 4H2O set up ratio based on molar masses (stat organized) 44g = 160g X 7.91g Solve for X I have found this method easier for students (gen. chem) as compared to dimensional analysis
