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Question 4.2: Relating the Numbers of Moles of Reactant and Product. How m......

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}?

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“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.


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}


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.

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