Question

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equation C 6 H 12 O 6 ( aq ) + 6 O 2 ( g ) ⟶ 6 CO 2 ( g ) + 6 H 2 O ( l ) Calculate the number of grams of oxygen required to convert 58.0 g of glucose to CO 2 and H 2 O .

Answer 1

Answer : The mass of
`O_2` required is, 61.82 grams.

Explanation :

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

`\text{Moles of }C_6H_(12)O_6=\frac{\text{Given mass }C_6H_(12)O_6}{\text{Molar mass }C_6H_(12)O_6}=(58.0g)/(180g/mol)=0.322mol`

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

The balanced chemical equation is:

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

From the balanced reaction we conclude that

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

So, 0.322 moles of
`C_6H_(12)O_6` react with
`0.322* 6=1.932` moles 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`

Molar mass of
`O_2` = 32 g/mole

`\text{ Mass of }O_2=(1.932moles)* (32g/mole)=61.82g`

Therefore, the mass of
`O_2` required is, 61.82 grams.