Glucose is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq) + 6O2(g) = 6CO2(g) + 6H2O(l). How many grams of oxygen are required to convert 28.0 g of glucose to CO2 and H2O?

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Glucose is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq) + 6O2(g) = 6CO2(g) + 6H2O(l). How many grams of oxygen are required to convert 28.0 g of glucose to CO2 and H2O?

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Answer 1 · 2017-10-15T00:05:58

28.1 grams to three significant figures.

Explanation:

Glucose $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ has a molecular mass of 180 grams.

$6 \times C = 6 \times 12 = 72$ $6 xx C = 6 xx 12 = 72$
$12 \times H = 12 \times 1 = 12$ $12 xx H = 12 xx 1 = 12$
$6 \times O = 6 \times 16 = 96$ $6 xx O = 6 xx 16 =96$ adding these gives
180 grams per mole.

To find the number of moles divide 28.0 grams by 180. grams

$\frac{28.0}{180} = .156$ $28.0/180 = .156$ moles

It requires 6 moles of Oxygen to "burn" 1 mole of glucose. so to find the number of moles of Oxygen required multiply by 6

$.156 \times 6 = .936$ $.156 xx 6 = .936$ moles of Oxygen required.

Oxygen has a molecular mass of 32

$2 \times O = 2 \times 16 = 32$ $2 xx O = 2 xx 16 = 32$ grams per mole.

To find the number of grams of Oxygen multiple moles by grams per mole.

$.936 \times 32 = 28.1$ $.936 xx 32 = 28.1$ grams Oxygen

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Glucose is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq) + 6O2(g) = 6CO2(g) + 6H2O(l). How many grams of oxygen are required to convert 28.0 g of glucose to CO2 and H2O?

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