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How many grams is 300 ml of butane?
170 to 190 grams depending on pressure. ie sea level or higher...
How much does a liter of butane weigh?
A liter of butane weighs approximately 0.58 kilograms (or 580 grams) at room temperature.
How many grams of Oxygen are needed to completely react with 5.0 grams of butane in a butane lighter?
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.
How many grams of CO2 is produced in the combustion of 100g butane?
C4H10 + 6.5 O2 -----> 4CO2 + 5H20 Mr of Butane = 58 5.8/58 = 0.1 So there is 0.1 moles of butane reacting 1 Mole = 24dm3 so 0.1 moles (5.8g of butane) = 0.1x24(dm3) = 2.4dm3 So then you see in the equation 4 moles of CO2 to 1 mole of C4H10 Thus 2.4 x 4 = Volume of CO2 = 9.6dm3 I think this is right, correct if wrong.
What is the mass of water vapor formed when 1g of butane is burned in a lighter?
Balanced equation. 2C4H10 + 13O2 -> 8CO2 +10H2O 8.13 grams C4H10 (1 mole C4H10/58.12 grams)(10 moles H2O/2 mole C4H10)(18.016 grams/1 mole H2O) = 12.6 grams water produced
How many grams is 300 ml of butane?
170 to 190 grams depending on pressure. ie sea level or higher...
What is butane density?
The density of butane is approximately 0.58 grams per cubic centimeter at 25°C (77°F) and atmospheric pressure.
What is the atomic mass of butane?
The atomic mass of butane (C4H10) is approximately 58.12 grams/mol.
How much does a liter of butane weigh?
A liter of butane weighs approximately 0.58 kilograms (or 580 grams) at room temperature.
How do you find the number of moles that will react with 5 grams of butane?
We need to know the number of moles of WHAT is to react with the butane to provide you with an answer.
How many grams of Oxygen are needed to completely react with 5.0 grams of butane in a butane lighter?
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.
How many grams of CO2 is produced in the combustion of 100g butane?
C4H10 + 6.5 O2 -----> 4CO2 + 5H20 Mr of Butane = 58 5.8/58 = 0.1 So there is 0.1 moles of butane reacting 1 Mole = 24dm3 so 0.1 moles (5.8g of butane) = 0.1x24(dm3) = 2.4dm3 So then you see in the equation 4 moles of CO2 to 1 mole of C4H10 Thus 2.4 x 4 = Volume of CO2 = 9.6dm3 I think this is right, correct if wrong.
What is the mass of water vapor formed when 1g of butane is burned in a lighter?
Balanced equation. 2C4H10 + 13O2 -> 8CO2 +10H2O 8.13 grams C4H10 (1 mole C4H10/58.12 grams)(10 moles H2O/2 mole C4H10)(18.016 grams/1 mole H2O) = 12.6 grams water produced
How many grams of oxygen are needed for the complete combustion of the butane in the lighter?
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.
How many grams of carbon dioxide would be produced by the complete combustion of 600 g of butane?
The balanced chemical equation for the combustion of butane is: 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O From the equation, 2 moles of butane produce 8 moles of carbon dioxide. So, 600 g of butane is approximately 7.23 moles. Therefore, the amount of carbon dioxide produced would be 29 moles, which is equivalent to 704 g.
What is the total mass of the carbon in 2 molecules of butane?
4 Carbon atoms in one molecule of Butane, times 2 because of two molecules, time the weight of one Carbon atom, which is 12 amu's (an amu = atomic mass unit, also known as a Dalton) equals 96 amu's.
How many grams of water are formed when 5.00g of butanol combust with excess oxygen?
To determine the number of grams of water formed, we need to calculate the moles of butanol (C4H9OH) and then use the balanced chemical equation to find the moles of water produced in the combustion reaction. From there, we convert moles of water to grams. The balanced equation for the combustion of butanol is C4H9OH + 6O2 → 4CO2 + 5H2O.
