Chapter 9, Problem 61E

Interpretation Introduction

Interpretation:

The mass of iron produced when $\text{50}\text{.0}\text{g}$ of molten iron $\left(\text{III}\right)$ oxide reacts with $\text{15}\text{.0}\text{g}$ of aluminum is to be calculated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, the reactant that controls the amount of the product formed is known as the limiting reactant.

Answer to Problem 61E

The mass of iron produced when $\text{50}\text{.0}\text{g}$ of molten iron $\left(\text{III}\right)$ oxide reacts with $\text{15}\text{.0}\text{g}$ of aluminum is $31.1\text{g}$.

Explanation of Solution

The reaction is given below.

${\text{Fe}}_2{\text{O}}_3\left(l\right)\text{+Al}\left(l\right)\stackrel{\Delta }{\to }\text{Fe}\left(l\right)+{\text{Al}}_2{\text{O}}_3\left(s\right)$

The balanced chemical equation for the reaction is given below.

${\text{Fe}}_2{\text{O}}_3\left(l\right)\text{+2Al}\left(l\right)\stackrel{\Delta }{\to }\text{2Fe}\left(l\right)+{\text{Al}}_2{\text{O}}_3\left(s\right)$

In the reaction, $1$ mole of ${\text{Fe}}_2{\text{O}}_3$ produces $2$ moles of $\text{Fe}$.

Therefore, the mole ratio is given below.

$\frac{1\text{mol}{\text{Fe}}_2{\text{O}}_3}{2\text{mol}\text{Fe}}$ and $\frac{2\text{mol}\text{Fe}}{1\text{mol}{\text{Fe}}_2{\text{O}}_3}$

The mole ratio to obtain moles of $\text{Fe}$ from moles of ${\text{Fe}}_2{\text{O}}_3$ is given below.

$\frac{2\text{mol}\text{Fe}}{1\text{mol}{\text{Fe}}_2{\text{O}}_3}$

The molar mass of iron is $55.85\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{Fe}}_2{\text{O}}_3$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times 55.85\text{g}{\text{mol}}^{-1}\right)+\left(3\times 16.00\text{g}{\text{mol}}^{-1}\right)\\ & =& 111.7\text{g}{\text{mol}}^{-1}+48.00\text{g}{\text{mol}}^{-1}\\ & =& 159.7\text{g}{\text{mol}}^{-1}\end{array}$

The formula to calculate the moles of ${\text{Fe}}_2{\text{O}}_3$ is given below.

$\text{Moles}\text{of}{\text{Fe}}_2{\text{O}}_3=\frac{\text{Mass}\text{of}{\text{Fe}}_2{\text{O}}_3}{\text{Molar}\text{mass}\text{of}{\text{Fe}}_2{\text{O}}_3}$ …(1)

The mass of ${\text{Fe}}_2{\text{O}}_3$ is $\text{50}\text{.0}\text{g}$.

Substitute the molar mass and mass of ${\text{Fe}}_2{\text{O}}_3$ in equation (1).

$\begin{array}{rll}\text{Moles}\text{of}{\text{Fe}}_2{\text{O}}_3& =& \frac{\text{50}\text{.0}\text{g}}{159.7\text{g}{\text{mol}}^{-1}}\\ & =& 0.313\text{mol}\end{array}$

The formula to calculate the moles of $\text{Fe}$ from moles of ${\text{Fe}}_2{\text{O}}_3$ is given below.

$\text{Moles}\text{of}\text{Fe}=\left(\begin{array}{l}\text{Moles}\text{of}{\text{Fe}}_2{\text{O}}_3\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}\text{Fe}\text{from}{\text{Fe}}_2{\text{O}}_3\end{array}\right)$ …(2)

Substitute the value of moles of ${\text{Fe}}_2{\text{O}}_3$ and mole ratio in equation (2).

$\begin{array}{rll}\text{Moles}\text{of}\text{Fe}& =& 0.313\text{mol}{\text{Fe}}_2{\text{O}}_3\times \frac{2\text{mol}\text{Fe}}{1\text{mol}{\text{Fe}}_2{\text{O}}_3}\\ & =& 0.626\text{mol}\end{array}$

In the reaction, $2$ moles of $\text{Al}$ produce $2$ moles of $\text{Fe}$.

Therefore, the mole ratio is given below.

$\frac{2\text{mol}\text{Al}}{2\text{mol}\text{Fe}}$ and $\frac{2\text{mol}\text{Fe}}{2\text{mol}\text{Al}}$

The mole ratio to obtain moles of $\text{Fe}$ from moles of $\text{Al}$ is given below.

$\frac{2\text{mol}\text{Fe}}{2\text{mol}\text{Al}}$

The molar mass of aluminum is $26.98\text{g}{\text{mol}}^{-1}$.

The formula to calculate the moles of $\text{Al}$ is given below.

$\text{Moles}\text{of}\text{Al}=\frac{\text{Mass}\text{of}\text{Al}}{\text{Molar}\text{mass}\text{of}\text{Al}}$ …(3)

The mass of $\text{Al}$ is $\text{15}\text{.0}\text{g}$.

Substitute the molar mass and mass of $\text{Al}$ in equation (3).

$\begin{array}{rll}\text{Moles}\text{of}\text{Al}& =& \frac{\text{15}\text{.0}\text{g}}{26.98\text{g}{\text{mol}}^{-1}}\\ & =& 0.556\text{mol}\end{array}$

The formula to calculate the moles of $\text{Fe}$ from moles of $\text{Al}$ is given below.

$\text{Moles}\text{of}\text{Fe}=\left(\begin{array}{l}\text{Moles}\text{of}\text{Al}\\ \times \text{Mole}\text{ratio}\text{to}\text{obtain}\text{moles}\text{of}\text{Fe}\text{from}\text{Al}\end{array}\right)$ …(4)

Substitute the value of moles of $\text{Al}$ and mole ratio in equation (4).

$\begin{array}{rll}\text{Moles}\text{of}\text{Fe}& =& 0.556\text{mol}\text{Al}\times \frac{2\text{mol}\text{Fe}}{2\text{mol}\text{Al}}\\ & =& 0.556\text{mol}\end{array}$

Since, $\text{15}\text{.0}\text{g}$ of $\text{Al}$ produces lesser amount of $\text{Fe}$, $\text{Al}$ is the limiting reactant.

The number of moles of $\text{Fe}$ produced by $\text{15}\text{.0}\text{g}$ of $\text{Al}$ is $0.556\text{mol}$.

The formula to calculate the mass of $\text{Fe}$ is given below.

$\text{Mass}\text{of}\text{Fe}=\text{Moles}\text{of}\text{Fe}\times \text{Molar}\text{mass}\text{of}\text{Fe}$ …(5)

The moles of $\text{Fe}$ is $0.556\text{mol}$.

The molar mass of $\text{Fe}$ is $55.85\text{g}{\text{mol}}^{-1}$.

Substitute the molar mass and moles of $\text{Fe}$ in equation (5).

$\begin{array}{rll}\text{Mass}\text{of}\text{Fe}& =& 0.556\text{mol}\times 55.85\text{g}{\text{mol}}^{-1}\\ & =& 31.1\text{g}\end{array}$

Therefore, the mass of iron produced when $\text{50}\text{.0}\text{g}$ of molten iron $\left(\text{III}\right)$ oxide reacts with $\text{15}\text{.0}\text{g}$ of aluminum is $31.1\text{g}$.

Conclusion

The mass of iron produced when $\text{50}\text{.0}\text{g}$ of molten iron $\left(\text{III}\right)$ oxide reacts with $\text{15}\text{.0}\text{g}$ of aluminum is $31.1\text{g}$.