Q. 7.8 - An Introduction to General, Organic, and Biological Chemistry

Question

Aluminum sulfate, ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$, is used in some antiperspirants.

1. How many moles of $\mathrm{O}$ are present in $3.0$ moles of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ ?
2.  How many moles of aluminum ions $\left({\mathrm{Al}}^{3+}\right)$ are present in $0.40$ mole of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ ?
3.  How many moles of sulfate ions $\left({{\mathrm{SO}}_4}^{2-}\right)$ are present in $1.5$ moles of ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ ? 

Step-by-Step Solution

Answer

Our required answers are:

1.  $36$moles of $\mathrm{O}$ atom
2. $0.80$moles of ${\mathrm{Al}}^{3+}$ ion
3.   $4.5$ moles of ${{\mathrm{SO}}_4}^{2-}$ ion

Part(a) Step 1: Given Information

We need to find moles of $\mathrm{O}$in$3$ moles of Aluminum sulfate .

Part(a) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$=  $12$ moles of  $\mathrm{O}$

Conversion Factor=$\frac{12 mole \mathrm{O}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of $\mathrm{O}$ in $3.00$ moles of  aluminum sulfate=$$\begin{gathered} \frac{12 mole \mathrm{O}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}\times 3 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =36 mole \mathrm{O} \end{gathered}$$

Hence,  $36$ moles of $\mathrm{O}$ present in $3.00$ moles aluminum sulfate.

Part(b) Step 1: Given Information

We need to find moles of ${\mathrm{Al}}^{3+}$ in $0.40$ moles of Aluminum sulfate .

Part(b) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$=  $2$ moles of  ${\mathrm{Al}}^{3+}$

Conversion Factor= $\frac{2 mole {\mathrm{Al}}^{3+}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of ${\mathrm{Al}}^{3+}$ ions in $0.40$ moles of aluminum sulfate= $$\begin{gathered} \frac{2 mole {\mathrm{Al}}^{3+}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}\times 0.40 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =0.80 mole {\mathrm{Al}}^{3+} \end{gathered}$$

Hence,  $0.80$ moles of ${\mathrm{Al}}^{3+}$ present in $0.40$ moles aluminum sulfate.

Part(c) Step 1: Given Information

We need to find moles of ${{\mathrm{SO}}_4}^{2-}$ ions  in $1.5$ moles of Aluminum sulfate .

Part(c) Step 2: Explanation

Using equalities and conversion factor using subscripts:

One mole of  ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$=  $3$ moles of   ${{\mathrm{SO}}_4}^{2-}$

Conversion Factor=  $\frac{3 mole {{\mathrm{SO}}_4}^{2-}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}$

Moles of  ions in  moles of aluminum sulfate=  $$\begin{gathered} \frac{3 mole {{\mathrm{SO}}_4}^{2-}}{1 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3}\times 1.5 mole {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3 \\ =4.5 mole {{\mathrm{SO}}_4}^{2-} \end{gathered}$$

Hence,  $4.5$ moles of ${{\mathrm{SO}}_4}^{2-}$ present in $1.5$ moles aluminum sulfate.