How many moles of co2 form when 58 grams of butane burn in oxygen?
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Asked in Elements and Compounds
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 grams of butane C4H10 burn in oxygen?
Asked in Chemical Equations
How many grams of o2 are needed to completely burn 96.6 grams of c3h8?
The chemical equation for this reaction is: C3H8(g) + 5O2(g) (arrow) 4H2O(l) + 3CO2(g) Using stoichiometry, 5 moles of oxygen are needed to burn 1 mole of C3H8. 96.6 grams is equal to 0.5889 moles. You multiply this by five to find the amount of moles of O2 required, which is 2.945 moles. Converting this to grams gives 94.24 grams of O2.
How many moles of oxygen are needed to burn 7.0 mols of phosphorus?
Asked in Chemistry
How many moles of O2 are needed to burn 1.50 moles of C8H18?
Asked in Chemistry
How many moles of O2 are needed to burn 10 moles of Mg?
Asked in Organic Chemistry
How many moles of oxygen required to burn 10 moles of ethene gas?
How many moles of O2 are needed to burn 1.35mol of C8H18?
Asked in Organic Chemistry, Elements and Compounds
Can propane be substituted for butane into a lighter?
Asked in Science, Chemistry, Stoichiometry
How many grams of 02(g) are needed to completely burn 23.7 g of C38H(g)?
How many moles of oxygen gas are needed to burn 10 moles of mg?
Asked in Hydrocarbons
How many moles of oxygen are required to burn 22.4 liters of ethane gas C2H6 at standard conditions?
Asked in Volume
How many liters of oxygen are needed to burn 20 liters of hydrogen?
Asked in Metal and Alloys
How many Grams of magnesium oxide are produced when 3 moles of magnesium burn?
A balanced equation for the reaction is 3 Mg + 3/2 O2 = 3 MgO. the atomic mass of mass of magnesium is about 24.3 and that of oxygen is about 16.0; therefore the mass of magnesium oxide produced is 3 (24.3 + 16.0) = 120.9 grams. (If the specification of "3 moles" of magnesium is considered to have only one significant digit, this answer should be written instead as 1 X 102 grams.)
Asked in Magnesium
What is the mass of magnesium oxide when you burn 7 grams of magnesium?
Mg grams -> (use Mg's molar mass) -> Mg moles -> (use ratio of moles - use balanced equation) -> MgO moles -> (use MgO's molar mass) -> grams MgO set up the equation: Mg + O2 --> MgO (we know the product is MgO and not MgO2 because magnesium has a charge of 2+ while oxygen has a charge or 2-) balance the equation: 2Mg + O2 --> 2MgO Molar mass of Mg: 24.31 g/mol Molar mass of MgO: 44.30 g/mol (add the molar mass of Magnesium - 24.31g/mol and the molar mass of Oxygen - 15.99g/mol together) (use periodic table to find these) 7.0 grams of Mg To find the moles of Magnesium you use the molar mass of Mg. (7.0 g Mg)*(1 mol Mg / 24.31 g Mg) =0.2879 moles Mg notice how the grams cancel to leave you with moles - remember dividing by a fraction is the same as multiplying by the reciprocal Now use the balanced equation's coefficients and the moles of Mg to determine the number of moles of MgO present. 2Mg + O2 --> 2MgO 2 moles Mg : 2 moles MgO -> divide both sides by 2 and it obviously becomes a 'one to one' ratio. This means that the number of moles of Mg is equal to the number of moles of MgO. This means that there are 0.2879 moles of MgO. Now that we know MgO's molar mass and the number of moles of MgO we have, the grams of MgO produced can be determined. (0.2879 moles MgO)*(44.30 g MgO / 1 mol MgO) = 12.75 grams MgO
Asked in Chemistry, Fossil Fuels
What volume of oxygen in liters would be required to burn one liter of gasoline petrol?
Asked in Elements and Compounds
What weight of oxygen is required to completely burn 19.8 g of octane?
The chemical equation for complete burning of octane is: 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O. This equation shows that 25 moles of diatomic oxygen are required to completely burn each two moles of octane. The gram molecular mass of octane is 114.23 and the gram molecular mass of diatomic oxygen is 2(15.9994). Therefore, the ratio of the mass of oxygen required to completely burn any given mass of octane to the mass of the octane to be burned is 50(15.9994)/2(114.23) or 3.5016, to the justified number of significant digits (the same as the number of digits in the least precisely specified datum 114.23, and the mass of oxygen required to burn 19.8 g of octane is (3.5016)(19.8) or 69.3 grams to the justified number of significant digits, now limited by the less precisely specified datum 119.8.