Chemistry · Ch 2 — Introduction to Analytical Chemistry

Limiting reagent

2.6

Limiting reagent

In a real laboratory reaction, the reactants are almost never supplied in the exact stoichiometric proportions given by the balanced equation. Because the point of running a reaction is usually to convert as much as possible of a costly or important starting material into the desired product, a chemist will often deliberately supply a large excess of one, cheaper reactant, precisely to make sure that the more expensive or more important reactant reacts completely. Whichever reactant is present in a stoichiometrically insufficient (lesser) amount gets used up first; once it is gone, the reaction simply stops, no matter how much of the other reactant is still sitting unreacted in the flask. Because this reactant is the one that runs out first and so limits — caps — the total amount of product the reaction can form, it is called the limiting reagent, and the reactant left over at the end is called the excess reagent. Identifying the limiting reagent is done by calculating, separately, how much product each reactant could form on its own if it reacted completely; whicheve …

Misc Worked ExampleIdentifying the limiting reagent in the formation of NO2 from NO and O2

Worked out. Worked example: nitrogen dioxide forms from nitric oxide and oxygen by $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$ . Starting with 8 moles of NO and 7 moles of $O_2$ , the limiting reagent is found by calculating how much $NO_2$ each reactant could produce on its own. From 8 mol NO: $8 \text{ mol NO} \times \dfrac{2 \text{ mol } NO_2}{2 \text{ mol NO}} = 8$ mol $NO_2$ . From 7 mol $O_2$ : $7 \text{ mol } O_2 \times \dfrac{2 \text{ mol } NO_2}{1 \text{ mol } O_2} = 14$ mol $NO_2$ . Since the limiting reagent is whichever one yields the SMALLER amount of product, and 8 moles NO give a smaller amount of $NO_2$ (8 mol) than 7 moles $O_2$ would (14 mol), NO is the limiting reagen …

Misc Problem 2.15Limiting reagent, product mass and leftover excess reagent in urea preparation

Worked out. Worked example: urea, $(NH_2)_2CO$ , is made by reacting ammonia with carbon dioxide, $2NH_3(g) + CO_2(g) \rightarrow (NH_2)_2CO(aq) + H_2O(l)$ . In one run, 637.2 g of $NH_3$ is treated with 1142 g of $CO_2$ . (a) Finding the limiting reagent: if all 637.2 g $NH_3$ reacted, moles of urea produced $= 637.2 \times \dfrac{1}{17.03} \times \dfrac{1}{2} = 18.71$ mol; if all 1142 g $CO_2$ reacted, moles of urea produced $= 1142 \times \dfrac{1}{44.01} \times \dfrac{1}{1} = 25.95$ mol. Since $NH_3$ gives the smaller amount of urea, $NH_3$ is the limiting reagent. (b) Mass of urea formed: using urea's molar mass of 60.06 g, mass $= 18.71 \times 60.06 = 1124$ g $(NH_2)_2CO$ . (c) Excess $CO_2$ remaining: the $CO_2$ actually consumed, from the mole ratio, is $18.71 \text{ mol} \times \dfrac{1 \text{ mol } CO_2}{1 \text{ mol urea}} \times 44.01 = 823.4$ g; t …

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