# 1.13 Reaction yields  (Page 2/9)

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$\text{mol HCl produced}=\text{3 mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{2 mol HCl}}{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}}\phantom{\rule{0.2em}{0ex}}=\text{6 mol HCl}$

Complete reaction of the provided chlorine would produce

$\text{mol HCl produced}=\text{2 mol}\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{2 mol HCl}}{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}}\phantom{\rule{0.2em}{0ex}}=\text{4 mol HCl}$

The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant ( [link] ).

## Identifying the limiting reactant

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

$\text{3Si}\left(s\right)+2{\text{N}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}{\text{N}}_{4}\left(s\right)$

Which is the limiting reactant when 2.00 g of Si and 1.50 g of N 2 react?

## Solution

Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

$\text{mol Si}=2.00\phantom{\rule{0.2em}{0ex}}\overline{)\text{g Si}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol Si}}{28.09\phantom{\rule{0.2em}{0ex}}\overline{)\text{g Si}}}\phantom{\rule{0.2em}{0ex}}=\text{0.0712 mol Si}$
$\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}=1.50\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}{28.02\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}}\phantom{\rule{0.2em}{0ex}}=\text{0.0535 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}$

The provided Si:N 2 molar ratio is:

$\frac{\text{0.0712 mol Si}}{\text{0.0535 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{1.33 mol Si}}{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}$

The stoichiometric Si:N 2 ratio is:

$\frac{\text{3 mol Si}}{\text{2 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{1.5 mol Si}}{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}$

Comparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.

Alternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

$\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}\phantom{\rule{0.2em}{0ex}}\text{produced}=\text{0.0712 mol Si}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}}{\text{3 mol Si}}\phantom{\rule{0.2em}{0ex}}=\text{0.0237 mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}$

while the 0.0535 moles of nitrogen would produce

$\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}\phantom{\rule{0.2em}{0ex}}\text{produced}=\text{0.0535 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}}{\text{2 mol}\phantom{\rule{0.2em}{0ex}}{\text{N}}_{2}}\phantom{\rule{0.2em}{0ex}}=\text{0.0268 mol}\phantom{\rule{0.2em}{0ex}}{\text{Si}}_{3}\phantom{\rule{0.1em}{0ex}}{\text{N}}_{4}$

Since silicon yields the lesser amount of product, it is the limiting reactant.

Which is the limiting reactant when 5.00 g of H 2 and 10.0 g of O 2 react and form water?

O 2

## Percent yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield    of the reaction. In practice, the amount of product obtained is called the actual yield    , and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield    :

$\text{percent yield}=\phantom{\rule{0.2em}{0ex}}\frac{\text{actual yield}}{\text{theoretical yield}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

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