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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 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 grams of butane C4H10 burn in oxygen?
The answer is 3,99 moles of carbon dioxide.
How many moles of co2 will be produced if 5.31 moles of c4h10 are used?
The combustion of butane (C₄H₁₀) can be represented by the balanced equation: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O. From this equation, we see that 2 moles of butane produce 8 moles of carbon dioxide. Therefore, if 5.31 moles of C₄H₁₀ are used, the moles of CO₂ produced can be calculated as follows: (5.31 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀) = 21.24 moles of CO₂.
How many hydrogen atoms are in a molecule of butane?
4
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 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 oxygen?
4 moles
How many moles of co2 form when 58 grams of butane burn in oxygen?
The answer is 3,99 moles of carbon dioxide.
How many moles of CO2 form when 58 grams of butane C4H10 burn in oxygen?
The answer is 3,99 moles of carbon dioxide.
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 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 sigma bonds are in a molecule of butane?
A molecule of butane contains a total of 10 sigma bonds.
How many grams of gaseous butane combust when 1550 kJ of heat are produced?
To determine the amount of butane that combusts, we need to use the enthalpy of combustion for butane, which is -2877 kJ/mol. Since the heat produced is 1550 kJ, we can set up a proportion to find the amount of butane consumed. By dividing the heat produced by the enthalpy of combustion per mole, we will get the number of moles of butane consumed. From there, you can convert moles to grams using the molar mass of butane (58.12 g/mol).
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.
