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This is an ignition.It forms CO2 and water.
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24g of CH4 react with 96 g of O2?
CH4 + 2O2 gives CO2 and 2H2O.So 16g of CH4 react with 64g of O2.So 24g react with 96g of O2
How do you balance the equation for CH4 O2 yield CO2 H2O?
Assuming complete combustion: CH4 + 2O2 --> 2H2O + CO2.
If 2.8 moles of CH4 react with 3 moles of O2 what is the limiting reagent?
Take the balanced equation. CH4+2O2---->CO2+2H2O.So 2.8mol of CH4 need 5.6mol of O2.So O2 is limitting factor
How do you balance the following equation ch4 plus o2 ------- co2 plus h20?
CH4+O2 --- CO2+H2O... All that's missing - is the number 2 before the water molecule... CH4+O2 --- CO2+2H2O
What is the balanced equation for methane burns oxygen to form water and carbon diooxide?
Unbalanced CH4 + O2 = H2O + CO2 Balanced CH4 + 2O2 = 2H20 + CO2
What is the balanced equation of CH4 plus O2 arrow CO2H2O?
The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O
When 85.0 g of CH4 are mixed with 160 g of O2 the limiting reactant is?
When 85.0 g of CH4 are mixed with 160. g of O2 the limiting reactant is __________. CH4 + 2O2 → CO2 + 2H2O
If 50 l of gaseous CH4 is burned at STP what volume of 02 is required for complete combustion and what volume of CO2 is produced?
CH4 + 2 O2 -> CO2 + 2 H2O So for every L of CH4 you need 2 L of O2 So for 50L of CH4 you need 100L O2
What do you get when you burn ch4 in the presence of o2?
It is a combusion.Water and CO2 is produced
How many molecules of oxygen 2 are needed to react with which molecule of methane CH4?
6 molecules of oxygen are needed to react with 3 methane molecules as one molecule of oxygen ( O2) are needed for methane gas.
Why does methane undergo oxidation?
Methane , CH4 , is a fuel that can react with O2 to yield CO2, H2O, and heat. CH4 (g) + O2 (g) ----> CO2 (g) + 2 H2O (g) + Heat
How many molecules of O2 ar needed in the chemical formula for burning methane?
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.
