How many molecules of O2 ar needed in the chemical formula for burning methane?

Answer 1

You should set up a balanced equation for the combustion of CH4 first:

CH4 + 2 O2 --> CO2 + 2 H2O

Now you see that for every 1 mole of CH4 you have 2 moles of O2. Therefore if you have 2.67 L of CH4, you'll have 5.34 L of O2.

You can also assume it's at STP just to check, since it all cancels out in the end:

2.67L CH4 (1mol CH4/22.4L)(2mol O2/ 1 mol CH4)(22.4L/1mol O2)= 5.34L O2.

Answer 2

Ethane is C2H6 and combustion of ethane is 2C2H6 + 7O2 ==> 4CO2 + 6H2OThus for combustion of 2 moles of ethane, one requires 7 moles O2

The mass of O2 required thus is 7 moles O2 x 32 g/mole O2 = 224 g of O2 needed

Answer 3

In air, methane is flammable over a fairly narrow range of concentrations, 5 to 15 percent. Thus there is a Lower Explosive Limit as well as a Higher Explosive Limit.

Answer 4

The equation for burning methane is: CH4 + 2 O2 => CO2 + 2 H2O. Therefore, two molecules of diatomic oxygen molecules are required to burn each molecule of methane.

Answer 5

The chemical formula for burning methane is CH4 + 2O2 → CO2 + 2H2O. This means that two molecules of O2 are needed for every one molecule of CH4 to completely burn methane.

Answer 6

0.500 l o2

Answer 7

25