Question

The thermite reaction has been used to weld railroad rails, make bombs, and ignite solid rocket fuel. The equation of the reaction is:



a.) What is the theoretical yield of iron if 100.0 g of Fe2O3 is used?

b.) What is the percentage yield if 52.3 g of iron is actually produced?

Answer 1

The thermite reaction has been used to weld railroad rails, make bombs, and ignite solid rocket fuel. The equation of the reaction is:

a.) What is the theoretical yield of iron if $100.0\ \mathrm{g}$ of $\mathrm{Fe}_2\mathrm{O}_3$ is used?

b.) What is the percentage yield if $52.3\ \mathrm{g}$ of iron is actually produced?

Solution:

The equation of the reaction is:

$$\mathrm{Fe}_2\mathrm{O}_3 + 2\mathrm{Al} = 2\mathrm{Fe} + \mathrm{Al}_2\mathrm{O}_3$$

Begin by determining the molar mass of each compound involved in the reaction.

Using atomic masses from the periodic table, we will find the following:

$$\begin{array}{l} \mathrm{M}(\mathrm{Fe}_2\mathrm{O}_3) = 159.7\ \mathrm{g/mol}; \\ \mathrm{Ar}(\mathrm{Fe}) = 55.85\ \mathrm{g/mol}. \end{array}$$

a) The theoretical yield of iron determine from the equation of the reaction:

$$\begin{array}{l} \mathrm{Fe}_2\mathrm{O}_3 + 2\mathrm{Al} = 2\mathrm{Fe} + \mathrm{Al}_2\mathrm{O}_3 \\ \mathrm{m}(\mathrm{Fe}) = \frac{100 \cdot 2 \cdot 55.85}{159.7} = 69.9\ \mathrm{g} \end{array}$$

b) The percentage yield of iron is:

$$\eta = \frac{52.3}{69.9} \cdot 100\% = 74.8\%.$$

Answer:

a) $\mathrm{m}(\mathrm{Fe}) = 69.9\ \mathrm{g}$;

b) yield of iron is $74.8\%$