# 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?

Oct 15, 2017

28.1 grams to three significant figures.

#### Explanation:

Glucose ${C}_{6} {H}_{12} {O}_{6}$ has a molecular mass of 180 grams.

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

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

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

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

Oxygen has a molecular mass of 32

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

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

$.936 \times 32 = 28.1$ grams Oxygen