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 $C_6H_{12}O_6$.

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

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 $0.294\times 6=1.764$ mole of $O_2$

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

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

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

Therefore, the mass of oxygen required are, 56.448 grams