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How many moles of oxygen are required to react with 5 moles of butane?
Assuming complete combustion of butane, you need 15 moles of oxygen to react with 5 moles of butane according to the balanced chemical equation: [ 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O ]
How many moles of CO2 form when 58.0 g of butane C4H10 burn in oxygen?
For the combustion of butane C4H10, the balanced chemical equation is: 2C4H10 + 13O2 -> 8CO2 + 10H2O. First, calculate the moles of butane: 58.0 g / 58.12 g/mol = 1 mole. From the balanced equation, 2 moles of butane produce 8 moles of CO2, so 1 mole of butane will produce 4 moles of CO2.
If 5.50 mol of butane are reacted how many moles of water vapor are produced?
Each mole of butane, which has formula of C4H10, contains 10 moles of hydrogen atoms. If the butane is completely combusted, all of the hydrogen in the butane is converted in water, with the formula H2O. The amount of water vapor will accordingly be 5.50 X 10/2 = 27.5.
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.
What is the maximum number of moles of carbon dioxide that could be produced when 105 grams of oxygen are reacted with excess of butane C4H10?
Assuming that the reaction is combustion, the balanced reaction equationwould be2 C4H10 +13 O2 ->8 CO2 +10 H2O. This shows that each mole of O2 would produce 8/13 mole of CO2. 105 grams of O2 corresponds to [105/(2 exact)(15.9994)] or about 3.281 moles of O2. or (8/13)(3.281) or 2.02 moles of carbon dioxide, to the justified number of significant digits.
How many grams of oxygen are required to burn 4.8 mol of butane?
The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.
How many moles of co2 form when 58.0 g of butane c4h10 burn oxygen?
4 moles
How many carbon atoms are in 2.00g of butane?
To find the number of carbon atoms in 2.00g of butane (C4H10), you first need to calculate the number of moles of butane using its molar mass (58.12 g/mol). Then, use Avogadro's number (6.022 × 10^23) to convert moles to atoms. Butane has 10 carbon atoms, so multiply the number of moles by 10 to find the number of carbon atoms.
How many moles of oxygen are required to react with 5 moles of butane?
Assuming complete combustion of butane, you need 15 moles of oxygen to react with 5 moles of butane according to the balanced chemical equation: [ 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O ]
How many moles of CO2 form when 58.0 g of butane C4H10 burn in oxygen?
For the combustion of butane C4H10, the balanced chemical equation is: 2C4H10 + 13O2 -> 8CO2 + 10H2O. First, calculate the moles of butane: 58.0 g / 58.12 g/mol = 1 mole. From the balanced equation, 2 moles of butane produce 8 moles of CO2, so 1 mole of butane will produce 4 moles of CO2.
How many moles of carbon dioxide (CO2) are produced when reacting 6.00 moles of butane (C4H10) in excess oxygen (O2)?
19,5 g butane are needed.
How many moles are there in fifteen kg of butane?
To find the number of moles in fifteen kg of butane, first calculate the molar mass of butane (C4H10) which is 58.12 g/mol. Then convert fifteen kg to grams (15000 g). Finally, divide the mass in grams by the molar mass to find the number of moles, which in this case is approximately 258.27 moles.
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 moles CO2 form when 58 g of butane C4H10 burn in oxygen?
To determine the moles of CO2 formed when 58 g of butane burns in oxygen, first, calculate the moles of butane using its molar mass. Then, use the stoichiometry of the balanced chemical equation to find the moles of CO2 formed, as per the ratio of the coefficients in the balanced equation.
How many moles of co2 form when 58 grams of butane burn in oxygen?
The answer is 3,99 moles of carbon dioxide.
What is the mass in grams of 3.00 moles of butane?
Butane is not a element. Butane has a molar mass.
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.
