The variables for the formula are incomplete. You would need to know how many grams of butane are put out by the lighter. The molecular weight of butane is 58.12 g/mol, which is also needed to complete the formula.
The balanced chemical equation for the combustion of butane is: 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O. This means it requires 13 moles of O₂ for every 2 moles of butane. The volume of O₂ needed would be (14.9 L * 13 mol) / 2 = 96.85 L of O₂.
To calculate the grams of oxygen needed, you first need to balance the chemical equation for the combustion of butane. C₄H₁₀ + O₂ → CO₂ + H₂O. From the balanced equation, 2 moles of butane react with 13 moles of oxygen. One mole of butane is 58.12 g, and one mole of oxygen is 32 g. Therefore, 5.0 g of butane would require (5.0 g / 58.12 g/mol) * 13 moles of oxygen, which is approximately 1.12 grams of oxygen.
Pure combustion of butane has the reaction 2 C4H10 + 13 O2 equals8 CO2 + 10 H2O. This means that every mole of butane produces 5 moles of water. 20 ml of water equals 1.11 moles, so .222 moles of butane are needed, or 5.17 liters.
The ideal air to fuel ratio for LPG (liquefied petroleum gas) combustion is typically around 25:1 by mass. This means 25 parts of air to 1 part of LPG for complete combustion. Adjustments may be needed based on specific engine requirements to achieve optimal performance and efficiency.
Oxygen is required to sustain combustion as it is the oxidizer that allows for the chemical reaction with the fuel. The exact amount of oxygen needed depends on the type of fuel being burned, as different fuels have different oxygen requirements for combustion. In general, combustion requires a sufficient supply of oxygen to ensure complete oxidation of the fuel.
The balanced chemical equation for the combustion of butane is: 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O. This means it requires 13 moles of O₂ for every 2 moles of butane. The volume of O₂ needed would be (14.9 L * 13 mol) / 2 = 96.85 L of O₂.
To calculate the grams of oxygen needed, you first need to balance the chemical equation for the combustion of butane. C₄H₁₀ + O₂ → CO₂ + H₂O. From the balanced equation, 2 moles of butane react with 13 moles of oxygen. One mole of butane is 58.12 g, and one mole of oxygen is 32 g. Therefore, 5.0 g of butane would require (5.0 g / 58.12 g/mol) * 13 moles of oxygen, which is approximately 1.12 grams of oxygen.
what is the mass in grams of oxygen, is needed to complete combustion of 6 L of methane?
9
1 mole
720 degrees
720 degrees
Pure combustion of butane has the reaction 2 C4H10 + 13 O2 equals8 CO2 + 10 H2O. This means that every mole of butane produces 5 moles of water. 20 ml of water equals 1.11 moles, so .222 moles of butane are needed, or 5.17 liters.
The ideal air to fuel ratio for LPG (liquefied petroleum gas) combustion is typically around 25:1 by mass. This means 25 parts of air to 1 part of LPG for complete combustion. Adjustments may be needed based on specific engine requirements to achieve optimal performance and efficiency.
Oxygen is required to sustain combustion as it is the oxidizer that allows for the chemical reaction with the fuel. The exact amount of oxygen needed depends on the type of fuel being burned, as different fuels have different oxygen requirements for combustion. In general, combustion requires a sufficient supply of oxygen to ensure complete oxidation of the fuel.
The balanced equation for complete combustion of heptane is C7H16 + 11O2 ---(ignition)---> 7CO2 + 8H2O. For each moelcule of heptane, you would need 11 molecules of oxygen gas.
The action is called "air-fuel ratio control" or "stoichiometric control." It ensures that the burner only allows in enough air to precisely match the amount of gas required for efficient and complete combustion.