Asked by Opo
calculate the number of oxygen needed to oxidize 12.5grams of glucose (C6H12O6) into carbon dioxide and water. (Combustion reaction)
Answers
Answered by
Jai
*i'll just repost my answer~ :)
first we write the combustion reaction:
C6H12O6 + O2 -> H2O + CO2
then we balance it:
C6H12O6 + 6O2 -> 6H2O + 6CO2
since mass of glucose is given, we get its molecular weight,, from the periodic table mass of each element in glucose is
H = 1 , C = 12 , O = 16
6*12 + 12*1 + 6*16 = 180 g/mol
thus,
12.5 g C6H12O6 * (1 mol C6H12O6 / 180 g C6H12O6) * (6 mol O2 / 1 mol C6H12O6) = 0.4167 mol O2
or in grams, (O2 = 32 g/mol)
32*0.4167 = 13.33 g O2
hope this helps~ :)
first we write the combustion reaction:
C6H12O6 + O2 -> H2O + CO2
then we balance it:
C6H12O6 + 6O2 -> 6H2O + 6CO2
since mass of glucose is given, we get its molecular weight,, from the periodic table mass of each element in glucose is
H = 1 , C = 12 , O = 16
6*12 + 12*1 + 6*16 = 180 g/mol
thus,
12.5 g C6H12O6 * (1 mol C6H12O6 / 180 g C6H12O6) * (6 mol O2 / 1 mol C6H12O6) = 0.4167 mol O2
or in grams, (O2 = 32 g/mol)
32*0.4167 = 13.33 g O2
hope this helps~ :)
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