## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 3.20

During strenuous exercise, respiration (the reaction between glucose and oxygen, Section 3.5) is limited by the availability of oxygen to the muscles. Under anaerobic conditions, our bodies produce lactic acid, which accumulates in muscles and accompanies the “burning” sensation you may have felt. We can model this situation by considering the reaction between 1.32 g of $C_{6}H_{12}O_{6}$ and 1.32 g of $O_{2}$ to produce $CO_{2}$ and water. Is one of these reactants a limiting reactant, or is the mixture stoichiometric? If one reactant is limiting, which one is it?

## Verified Solution

Collect and Organize We are given quantities of two reactants $(C_{6}H_{12}O_{6} and CO_{2})$ and want to determine which, if either, is limiting.

Analyze We can determine the limiting reactant by calculating the ratio of $C_{6}H_{12}O_{6}$ to $O_{2}$ in the balanced chemical equation for the reaction. That means we will use the second method described in the preceding discussion. An unbalanced equation for this combustion reaction is

$C_{6}H_{12}O_{6}(s)+O_{2}(g)→CO_{2}(g)+H_{2}O(g)$

After balancing this equation, we compare the ratio of the amounts of reactants with the stoichiometric ratio in the balanced equation to determine whether one of the reactants is limiting.

Solve First we need to balance the combustion equation:

_$C_{6}H_{12}O_{6}(s)+$_$O_{2}(g) →$_$CO_{2}(g)+$_$H_{2}O(g)$

$C_{6}H_{12}O_{6}(s)+6 O_{2}(g) → 6 CO_{2}(g)+6H_{2}O(g)$

This balanced equation tells us that the mole ratio of $C_{6}H_{12}O_{6}$ to $O_{2}$ is 1:6. Next we convert the given masses of reactants to moles:

$1.32 \sout{g O_{2}} \times \frac{1 mol O_{2}}{32.00 \sout{g O_{2}}} =4.13 \times 10^{-2} mol O_{2}$

$1.32 \sout{g C_{6}H_{12}O_{6}} \times \frac{1 mol C_{6}H_{12}O_{6}}{180.16 \sout{g C_{6}H_{12}O_{6}}}=7.33 \times 10^{-3} mol C_{6}H_{12}O_{6}$

Then we calculate the mole ratio:

$\frac{4.13 \times 10^{-2} mol O_{2}}{7.33 \times 10^{-3} mol C_{6}H_{12}O_{6}}=5.6$

This ratio is less than the stoichiometric $C_{6}H_{12}O_{6}$ to $O_{2}$ of 1:6, so oxygen is the limiting reactant, and there is a slight excess of glucose.

Think About It The calculation indicates that when equal masses of glucose and oxygen react, oxygen is the limiting reactant.