Q. 4.2

Relating the Numbers of Moles of Reactant and Product

How many moles of $CO_{2}$ are produced in the combustion of 2.72 mol of triethylene glycol, $C_{6}H_{14}O_{4}$, in an excess of $O_{2}$?

Verified Solution

Analyze

“An excess of  $O_{2}$” means that there is more than enough $O_{2}$ available to permit the complete conversion of the triethylene glycol to $CO_{2}$ and $H_{2}O$. The factor for converting from moles of $C_{6}H_{14}O_{4}$ to moles of $CO_{2}$ is obtained from the balanced equation for the combustion reaction.

Solve

The first step in a stoichiometric calculation is to write a balanced equation for the reaction. The balanced chemical equation for the reaction is given below.

2 $C_{6}H_{14}O_{4} + 15 O_{2} → 12 CO_{2} + 14 H_{2}O$

Thus, 12 mol $CO_{2}$ are produced for every 2 mol $C_{6}H_{14}O_{4}$ burned. The production of 12 mol $CO_{2}$ is equivalent to the consumption of 2 mol $C_{6}H_{14}O_{4}$; thus, the ratio 12 mol $CO_{2}$/2 mol $C_{6}H_{14}O_{4}$ converts from mol $C_{6}H_{14}O_{4}$ to mol $CO_{2}$.

? mol $CO_{2} = 2.72 mol C_{6}H_{14}O_{4} × \frac{12 mol CO_{2}}{2 mol C_{6}H_{14}O_{4}} = 16.3 mol CO_{2}$

Assess

The expression above can be written in terms of two equal ratios:

$\frac{? mol CO_{2}}{2.72 mol C_{6}H_{14}O_{4}} = \frac{12 mol CO_{2}}{2 mol C_{6}H_{14}O_{4}}$

You may find it easier to set up an expression in terms of ratios and then solve it for the unknown quantity.