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 $C_6H_{12}O_6$

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

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 $0.322\times 6=1.932$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

Molar mass of $O_2$ = 32 g/mole

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

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