Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Percent Yield from Chemical Reactions - Page 109: 48

a) $350g$ b) $47.57\%$

Let's consider the reaction that takes place: $Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$ Since we are given the mass of both reagents, first we need to determine which one is the limiting reagent. We can calculate the amounts of both of reagents: $n(Fe_2O_3) = \frac{m(Fe_2O_3)}{M(Fe_2O_3)} = \frac{500g}{160\frac{g}{mol}} = 3.125mol$ $n(Al) = \frac{m(Al)}{A(Al)} = \frac{500g}{27\frac{g}{mol}} = 18.52mol$ Observing stoichiometric coefficients, we can conclude that the amount of aluminum required for the reaction is twice the amount of iron(III) oxide, which means that for the reaction of $3.125mol$ of $Fe_2O_3$, $6.25mol$ of $Al$ is required. Therefore, iron(III) oxide is the limiting reagent. a) The amount of iron produced in this reaction is twice the amount of iron(III) oxide, hence the amount of the obtained iron is equal to $6.25mol$, in theory. That corresponds to the following mass of iron: $m(Fe) = n(Fe)\times A(Fe) = 6.25mol \times 56\frac{g}{mol} = 350g$ b) The percent yield of this reaction can be obtained by diving the actual mass of produced iron by the theoretical mass obtained in this reaction. Hence, the percent yield of this reaction is $\frac{166.5g}{350g} \approx 47.57\%$.