Explanation:
(a) When oxidation of glucose occurs the glucose molecules react with oxygen and it results in the formation of carbon dioxide and water.
And, when in a chemical reaction equation number of atoms on the reactant side are equal to the number of atoms on product side then it is known as a balanced equation.
For example, [tex]C_{6}H_{12}O_{6} + O_{2} \rightarrow CO_{2} + H_{2}O[/tex]
Number of atoms on reactant side are as follows.
C = 6
H = 12
O = 8
Number of atoms on product side are as follows.
C = 1
H = 2
O = 3
Therefore, balance this equation by multiplying [tex]O_{2}[/tex] by 6 on reactant side. Also, on reactant side multiply [tex]CO_{2}[/tex] by 6 and [tex]H_{2}O[/tex] by 6.
Hence, the balanced equation is as follows.
[tex]C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O[/tex]
(b) According to the reaction equation, to completely oxidize 1 mole of glucose we need 6 mole of [tex]O_{2}[/tex].
Therefore, to completely oxidize 180 g of glucose we need, [tex]6 \times 32 g[/tex] of [tex]O_{2}[/tex].
So, calculate the the amount of [tex]O_{2}[/tex] required to completely oxidize 1 g of glucose as follows.
[tex]\frac{6 \times 32 g}{180 g}[/tex] g of [tex]O_{2}[/tex]
= 1.07 g
Thus, we can conclude that the mass of [tex]O_{2}[/tex] required to completely oxidize 1 g of glucose is 1.07 g.