CH4 + 2O2 gives CO2 and 2H2O.So 16g of CH4 react with 64g of O2.So 24g react with 96g of O2
Unbalanced CH4 + O2 = H2O + CO2 Balanced CH4 + 2O2 = 2H20 + CO2
O2
The combustion of methane can be balanced in the following manner. One molecule of CH4 plus two molecules of O2 produces one molecule of CO2 plus one molecule of H2O.
I assume you were looking to balance the reaction: CH4 + 2O2 --> CO2 +2H2O
1.5 moles of O2 react with 2 moles of Al 2Al + 1.5 O2 = Al2O3.
Take the balanced equation. CH4+2O2---->CO2+2H2O.So 2.8mol of CH4 need 5.6mol of O2.So O2 is limitting factor
6 molecules of oxygen are needed to react with 3 methane molecules as one molecule of oxygen ( O2) are needed for methane gas.
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
Take the balanced equation CH4+2O2---->CO2+2H2O.So according to it 40g of O2
Assuming complete combustion: CH4 + 2O2 --> 2H2O + CO2.
128 g of oxygen are needed.
CH4+O2 --- CO2+H2O... All that's missing - is the number 2 before the water molecule... CH4+O2 --- CO2+2H2O
The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O
Unbalanced CH4 + O2 = H2O + CO2 Balanced CH4 + 2O2 = 2H20 + CO2
When 85.0 g of CH4 are mixed with 160. g of O2 the limiting reactant is __________. CH4 + 2O2 → CO2 + 2H2O
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
CH4 + 2O2 → CO2 + 2H2O Methane + Oxygen → Carbon dioxide + Water