You can approach this from a stoichiometry perspective or a proportions perspective.
Before you do one of these methods, you want to calculate the molar mass of ethane. Ethane has two carbons and six hydrogens. Using the Periodic Table, we find that the molar mass is 30.06 g/mol.
If you do the stoichiometry, we have (1 mol)×(30.06 g/mol)×(312 kJ / 6.0 g) = 1600 kJ.
Or you can say that if 6.0 g = 312 kJ, then 30.06 g = x kJ where "x" is your unknown, or (6 g / 312 kJ) = (30.06 g / x kJ). Solving for x provides 1600 kJ.
Both involve the same calculations but have slightly different modes of thinking. Note that the answer has only two significant digits because 6.0 has only two significant digits. We can assume that 1 mole is a discrete value and that our answer shouldn't have one significant digit.
Combustion.
The reduction of iodomethane with a mixture of Zinc and hydrochloric acid produces ethane.
Ethane does not have any molecule of carbon dioxide. However when ethane undergoes combustion then two molecules of carbon dioxide are formed (as ethane contains two carbon atoms).
Yes. Ethane can undergo combustion, in which it reacts with oxygen to produce carbon dioxide and water. 2C2H6 + 7O2 --> 4CO2 + 6H2O
Yes. Ethane can undergo combustion, in which it reacts with oxygen to produce carbon dioxide and water. 2C2H6 + 7O2 --> 4CO2 + 6H2O
-51.88 kJ/g
Combustion.
The reduction of iodomethane with a mixture of Zinc and hydrochloric acid produces ethane.
Ethane does not have any molecule of carbon dioxide. However when ethane undergoes combustion then two molecules of carbon dioxide are formed (as ethane contains two carbon atoms).
Yes there are. These gases are hydrogen,methane,ethane,propane,butane,pentane. Hope this helped! ;)
Yes. Ethane can undergo combustion, in which it reacts with oxygen to produce carbon dioxide and water. 2C2H6 + 7O2 --> 4CO2 + 6H2O
Yes. Ethane can undergo combustion, in which it reacts with oxygen to produce carbon dioxide and water. 2C2H6 + 7O2 --> 4CO2 + 6H2O
Reaction of combustion of methane will give off lot of energy. In any combustion reaction there will always be formation of water vapor and heat. Methane + oxygen = combustion reaction.
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Combustion of Ethane: 2C2H6+7O2-->4CO2+6H2O Combustion of Ethanol: C2H5OH+3O2-->2CO2+3H2O
2C2H6 + 7O2 --> 4CO2 + 6H2O You must have the same total number of atoms of each element on either side of the arrow. To do this, you must use whole-number coefficients (as needed) in front of each reactant and product. They must be expressed in the lowest possible ratio.
H. T. Lai has written: 'Numerical study of contaminant effects on combusstion if hydrogen,ethaneand methane in air' -- subject(s): Hydrogen fuels, Carbon dioxide, Nitrous oxides, Water, Ethane, Rection kinetics, Contaminants, Methane, Combustion chemistry 'Numerical study of contaminant effects on combustion of hydrogen, ethane and methane in air'