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What is the balanced equation of CH4 plus O2 arrow CO2H2O?
The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O
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
If 2.8 moles of CH4 react with 3 moles of O2 what is the limiting reagent?
To determine the limiting reagent, we need to compare the amount of each reactant to their stoichiometric coefficients in the balanced equation. The balanced equation for the combustion of CH4 with O2 is: CH4 + 2O2 -> CO2 + 2H2O. From the given amounts, we can see that O2 is in excess, so CH4 is the limiting reagent.
What type of reaction is CH4 plus 2O2 equals CO2 plus 2H2O plus 218kcal?
The given equation represents a combustion reaction, where methane (CH4) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). The release of 218 kcal indicates that this reaction is exothermic, releasing heat energy.
Molecular equation for burning methane gas?
The molecular equation for the combustion of methane gas (CH4) in the presence of oxygen (O2) is: CH4 + 2O2 -> CO2 + 2H2O.
What is the balanced equation of CH4 plus O2 arrow CO2H2O?
The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O
Balance this equation CH4 plus O2 equals CO2 plus H2O?
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.
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
CH4 plus O2 equals H20 plus?
The chemical reaction is: CH4 + 2O2 --> 2H2O + CO2 CO2 is the carbon dioxide.
What type of chemical reaction is CH4 plus 2 O2 -- CO2 plus 2 H2O?
Burning, oxidation.
If 2.8 moles of CH4 react with 3 moles of O2 what is the limiting reagent?
To determine the limiting reagent, we need to compare the amount of each reactant to their stoichiometric coefficients in the balanced equation. The balanced equation for the combustion of CH4 with O2 is: CH4 + 2O2 -> CO2 + 2H2O. From the given amounts, we can see that O2 is in excess, so CH4 is the limiting reagent.
How do you balance the equation for CH4 O2 yield CO2 H2O?
The balanced chemical equation for the reaction CH4 + 2O2 → CO2 + 2H2O shows that 1 molecule of CH4 reacts with 2 molecules of O2 to produce 1 molecule of CO2 and 2 molecules of H2O. This equation ensures that the number of atoms of each element is the same on both sides of the reaction arrow.
CH4 plus O2 CO2 plus H2O?
I assume you were looking to balance the reaction: CH4 + 2O2 --> CO2 +2H2O
What type of reaction is CH4 plus 2O2 equals CO2 plus 2H2O plus 218kcal?
The given equation represents a combustion reaction, where methane (CH4) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). The release of 218 kcal indicates that this reaction is exothermic, releasing heat energy.
What is the balance equation of methane burned in air?
The balanced chemical equation for methane (CH4) burning in air (O2) is: CH4 + 2O2 → CO2 + 2H2O
When 3.05 of ch4 is mixed with 5.03 moles of o2 the limiting reactants is?
The balanced chemical equation for the reaction between CH4 and O2 is as follows: CH4 + 2O2 -> CO2 + 2H2O By comparing the moles of CH4 and O2 given, you can see that O2 is the limiting reactant because it will be completely consumed before all of the CH4 can react.
What mass of oxygen is required to completely react with 20 grams of CH4?
To calculate the mass of oxygen required to react with 20 grams of CH4, we first need to write and balance the chemical equation for the reaction. The balanced equation for the combustion of CH4 is: CH4 + 2O2 → CO2 + 2H2O This equation tells us that 1 mole of CH4 reacts with 2 moles of O2. The molar mass of CH4 is 16 g/mol. Therefore, 20 grams of CH4 is equal to 20/16 = 1.25 moles CH4. So, 1.25 moles of CH4 would require 2.50 moles of O2. The molar mass of O2 is 32 g/mol. Therefore, the mass of O2 required would be 2.50 moles * 32 g/mol = 80 grams.
