Question

Glucose, C6H12O6, 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)

Calculate the number of grams of oxygen required to convert 53.0 g of glucose to CO2 and H2O.

Answer 1

Answer : The mass of oxygen required are, 56.448 grams

Explanation : Given,

Mass of glucose = 53 g

Molar mass of glucose = 180.156 g/mole

Molar mass of
`O_2` = 32 g/mole

First we have to calculate the moles of
`C_6H_(12)O_6`.

`\text{Moles of }C_6H_(12)O_6=\frac{\text{Mass of }C_6H_(12)O_6}{\text{Molar mass of }C_6H_(12)O_6}=(53.0g)/(180.156g/mole)=0.294moles`

Now we have to calculate the moles of
`O_2`.

The balanced chemical reaction will be,

`C_6H_(12)O_6(aq)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l)`

From the balanced chemical reaction, we conclude that

As, 1 mole of
`C_6H_(12)O_6` react with 6 moles of
`O_2`

So, 0.294 mole of
`C_6H_(12)O_6` react with
`0.294* 6=1.764` mole of
`O_2`

Now we have to calculate the mass of
`O_2`.

`\text{Mass of }O_2=\text{Moles of }O_2* \text{Molar mass of }O_2`

`\text{Mass of }O_2=(1.764mole)* (32g/mole)=56.448g`

Therefore, the mass of oxygen required are, 56.448 grams