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The combustion of glucose is represented by the following balanced equation:

The combustion of glucose is represented by the following balanced equation:-example-1

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In order to find limiting and excess reactants, we need to first, set up the equation, and it must be properly balanced:

C6H12O6 + 6 O2 -> 6 H2O + 6 CO2

1 gram of C6H12O6

1 gram of O2

From the balanced equation, we see that the molar ratio between C6H12O6 and O2 is 1:6, this means that 1 mol of C6H12O6 will react with 6 times the number of moles of O2

Now let's find the number of moles of C6H12O6, the molar mass is 180.16g/mol

180.16 g = 1 mol

1 g = x moles

x = 0.005 moles of C6H12O6, now we will need this number of moles times 6 in order to proceed with the reaction

1 C6H12O6 = 6 O2

0.005 C6H12O6 = x O2

x = 0.03 moles of O2

And this amount of moles with the molar mass of O2, which is 32g/mol

32 g = 1 mol

x grams = 0.03 moles

x = 0.96 grams of O2 in 0.03 moles

Since we have more oxygen than what we actually need, Oxygen is in excess and C6H12O6 is the limiting reactant

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