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?

1 Answer
Oct 15, 2017

28.1 grams to three significant figures.

Explanation:

Glucose C_6H_12O_6C6H12O6 has a molecular mass of 180 grams.

6 xx C = 6 xx 12 = 726×C=6×12=72
12 xx H = 12 xx 1 = 1212×H=12×1=12
6 xx O = 6 xx 16 =966×O=6×16=96 adding these gives
180 grams per mole.

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

28.0/180 = .156 28.0180=.156moles

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 xx 6 = .936.156×6=.936 moles of Oxygen required.

Oxygen has a molecular mass of 32

2 xx O = 2 xx 16 = 32 2×O=2×16=32 grams per mole.

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

.936 xx 32 = 28.1 .936×32=28.1 grams Oxygen