Chapter 23, Problem 87A

Interpretation Introduction

Interpretation:

The mass of glucose in grams should be calculated which can be oxidised completely by 2.0 L of oxygen gas at STP during cellular respiration.

Concept introduction:

A set of metabolic reactions which takes place in the organism cell to convert chemical energy from molecules of oxygen into ATP that is adenosine triphosphate is known as cellular respiration. In this process, glucose breaks down in the presence of oxygen and produces carbon dioxide with release of water and energy.

The chemical reaction is:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{+6O}}_{\text{2}}\to {\text{6CO}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O+energy}$

Answer to Problem 87A

Mass of glucose in grams = 2.7 g.

Explanation of Solution

Volume = 2.0 L

The chemical reaction for the cellular respiration process is:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{+6O}}_{\text{2}}\to {\text{6CO}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O+energy}$

According to the above reaction, one mole of glucose reacts with six moles of oxygen to form six moles of carbon dioxide and six moles of water.

Since, at STP:

One mole of any gas occupies volume 22.4 L.

Thus, number of moles of oxygen = $2.0\text{ L}\times \frac{1\text{ mole}}{22.4{\text{ L O}}_2}$

= $0.0893\text{ mole}$

According to the reaction, the ratio between glucose and oxygen is 1:6.

Thus, number of moles of glucose = $\frac{0.089\text{ mole}}{6}$

= $0.01488\text{ mole}$

Molar mass of glucose = 180.16 g/mole

Number of moles = $\frac{\text{mass}}{\text{molar mass}}$

Put the values,

$0.01488\text{ mole}$ =

$\frac{\text{mass}}{\text{180}\text{.16 g/mole}}$

Mass of glucose = $0.01488\text{ mole}\times \text{180}\text{.16 g/mole}$

= 2.7 g

Thus, mass of glucose in grams = 2.7 g