Question 3.20: During strenuous exercise, respiration (the reaction between...

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.