Question

Over the years, the thermite reaction has been used for welding railroad rails, in incendiarybombs, and to ignite solid-fuel rocket motors. The reaction is

${\mathbf{\text{Fe}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}\left(\text{s}\right)\mathbf{\text{+2Al}}\left(\text{s}\right)\mathbf{\text{→2Fe}}\left(\text{l}\right){\mathbf{\text{+Al}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}\left(s\right)$

what masses of iron (III) oxide and aluminum must be used to produce $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathit{g}$ of iron? What isthe maximum mass of aluminum oxide that could be produced?

Short answer

The masses of iron (III) oxide and aluminum must be used to produce$15.0g$ are role="math" localid="1648554037572" $\mathbf{21}\mathbf{.}\mathbf{4}\mathbf{}\mathit{g}$and $\mathbf{7}\mathbf{.}\mathbf{24}\mathbf{}\mathit{g}$, and the maximum mass of aluminum oxide that can be produced is$\mathbf{13}\mathbf{.}\mathbf{7}\mathbf{}\mathit{g}$ .

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determining the moles of 15.8 g of iron  Fe2O3

Given balanced thermite reaction as follows:

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

The moles of $15.0g$of iron = role="math" localid="1648552195439" $\frac{\mathbf{\text{Mass}}}{\mathbf{\text{AtomicMass}}}$

$$\begin{gathered} =\frac{15.0}{56} \\ =0.268\text{Moles} \end{gathered}$$

.

Determining the mass of iron (iii) oxide  

For 2 moles of iron, 1 mol${\text{Fe}}_2{\text{O}}_3$is needed.

For $0.268$ moles of iron$=\frac{1}{2}\times 0.268$

Molar mass of${\text{Fe}}_2{\text{O}}_3$

role="math" localid="1648551504832" $$\begin{gathered} \text{=2}\left(\text{Fe}\right)\text{+3}\left(\text{O}\right) \\ \text{=2}\left(56\right)\text{+3}\left(16\right) \\ \text{=169}\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right. \end{gathered}$$

Mass of iron (iii) oxide $=\left(\text{moles}\right)\times \left(molarmass\right)$

$=\left(0.134mol\right)\times \left(160\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.\right)$

Therefore, the mass of iron (iii) oxide is$\mathbf{21}\mathbf{.}\mathbf{4}\mathbf{}\mathit{g}$ of ${\mathbf{\text{Fe}}}_{\mathbf{2}}{\mathbf{\text{O}}}_{\mathbf{3}}$

Identification of produced Aluminum

From the given equation, for 2 moles of Aluminum, 2 mol of $\text{Fe}$

So, $0.268$ moles of iron will be produced from $0.268$ moles of Aluminum

Now, the mass of $0.268$ moles of Aluminum role="math" localid="1648554221738" $\mathbf{=}\frac{\mathbf{\text{Mass}}}{\mathbf{\text{Atomicmass}}}$

$=\left(0.268\right)\times \left(27\right)$

Therefore, the mass of $0.268$ moles of Aluminum is role="math" localid="1648551049314" $\mathbf{7}\mathbf{.}\mathbf{24}\mathbf{}\mathit{g}\mathbf{}$of role="math" localid="1648551059271" $\mathbf{\text{Al}}$.

Determining maximum mass of Aluminum oxide that could be produced

2 moles of Aluminum can produce 1 mol of ${\text{Al}}_2{\text{O}}_3$

Thus $0.268$moles of Aluminum can produce $=\frac{0.268}{2}$

$=0.134$Moles of ${\text{Al}}_2{\text{O}}_3$

Molar mass $=2\left(27\right)+3\left(16\right)$

$=102\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Thus the mass of ${\text{Al}}_2{\text{O}}_3\text{=}\left(0.134\right)\text{×}\left(102\right)$

$=13.7g$

Therefore the maximum of ${\text{Al}}_2{\text{O}}_3$that can be produced isrole="math" localid="1648550601767" $\mathbf{13}\mathbf{.}\mathbf{7}\mathbf{}\mathit{g}$ .