Question

How many grams of oxygen does it take to burn 16.8 g of glucose?

Answer 1

The combustion reaction of glucose is as follows:

C₆H₁₂O₆ + 6O₂ 6CO₂ + 6H₂O

Molar mass of C₆H₁₂O₆ = (6xAt.mass of C) + (12 xAt.mass of H) + (6xAt.mass of O)

                                    = ( 6x12) + (12x1) + (6x16)

                                    = 180 g/mol

Molar mass of O₂ = 2xAt.mass of O

                          = 2x16

                         = 32 g/mol

According to the balanced equation ,

1 mole of C₆H₁₂O₆ reacts with 6 moles of O₂

                           OR

1 x180 g of C₆H₁₂O₆ reacts with 6x32 g of O₂

16.8 g of C₆H₁₂O₆ reacts with M g of O₂

M = (16.8x6x32) / (1x180)

   = 17.9 g of oxygen

Therefore the mass of oxygen required is 17.9 g