2.4 x (6.022x1023) = 1.44528x1024 molecules of butane
number of moles (of molecules) x avgadro's constant(6.022x1023) = number of molecules
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.
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.
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 ]
The answer is 3,99 moles of carbon dioxide.
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.
4
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.
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.
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 ]
4 moles
The answer is 3,99 moles of carbon dioxide.
The answer is 3,99 moles of carbon dioxide.
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.
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.
A molecule of butane contains a total of 10 sigma bonds.
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).
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.