Calculating the Mass of Al2(SO4)3 in a Sodium Aluminate Solution

In this article, we will delve into the detailed calculation of the mass of aluminum sulfate, Al2(SO4)3, in a sodium aluminate solution. The given solution has a normality (N) of 0.48 N in a 250 ml volume. We will explore four different methods to arrive at the final answer, ensuring clarity and accuracy in the steps involved.

Method 1: Using Molar Mass and Equivalent Weight

Step 1: Calculate the Molar Mass of Al2(SO4)3

The molar mass of Al2(SO4)3 is calculated as follows:

Molar mass 2 × 27.0 g/mol (Al) 3 × 32.1 g/mol (S) 12 × 16.0 g/mol (O)

Molar mass 54.0 96.3 192.0 342.3 g/mol

Step 2: Calculate the Equivalent Weight

The equivalent weight of Al2(SO4)3 is calculated as:

Equivalent weight Molar mass / Number of equivalents per mole

For Al2(SO4)3, there are 2 Al3 ions and 3 SO42- ions, so the number of equivalents 2 3 6

Equivalent weight 342.3 g/mol / 6 57.05 g/eq

Step 3: Calculate the Number of Equivalents in the Solution

No. of equivalents of Al2(SO4)3 in the solution Normality × Volume / 1000

No. of equivalents of Al2(SO4)3 0.48 eq/L × 250 ml / 1000

No. of equivalents of Al2(SO4)3 0.12 eq

Step 4: Calculate the Mass of Al2(SO4)3 in the Solution

Mass of Al2(SO4)3 No. of equivalents × Equivalent weight

Mass of Al2(SO4)3 0.12 eq × 57.05 g/eq 6.85 g

Method 2: Using Concentration and Volume

Step 1: Calculate the Concentration of Al2(SO4)3

The concentration of Al2(SO4)3 is calculated as:

Concentration Normality * 1 M / 6 N

Concentration 0.48 M / 6 0.08 M

Step 2: Calculate the Number of Moles of Al2(SO4)3 in the Solution

No. of moles of Al2(SO4)3 Concentration × Volume / 1000

No. of moles of Al2(SO4)3 0.08 mol/L × 250 ml / 1000

No. of moles of Al2(SO4)3 0.02 mol

Step 3: Calculate the Mass of Al2(SO4)3 in the Solution

Mass of Al2(SO4)3 No. of moles × Molar mass

Mass of Al2(SO4)3 0.02 mol × 342.3 g/mol 6.85 g

Method 3: Using Equivalents, Volume, and Equivalent Weight

No. of equivalents of Al2(SO4)3 (0.48 eq/L × 250 ml) / 1000

No. of equivalents of Al2(SO4)3 0.12 eq

Mass of Al2(SO4)3 No. of equivalents × Equivalent weight

Mass of Al2(SO4)3 0.12 eq × 57.05 g/eq 6.85 g

Method 4: Using Concentration, Volume, Moles, and Molar Mass

No. of equivalents of Al2(SO4)3 (0.48 eq/L × 250 ml) / 1000

No. of equivalents of Al2(SO4)3 0.12 eq

No. of moles of Al2(SO4)3 No. of equivalents / 6

No. of moles of Al2(SO4)3 0.12 / 6 0.02 mol

Mass of Al2(SO4)3 No. of moles × Molar mass

Mass of Al2(SO4)3 0.02 mol × 342.3 g/mol 6.85 g

Conclusion

Using the four different methods described above, we have arrived at the same result: the mass of Al2(SO4)3 in the solution is 6.85 g. This method not only provides a deeper understanding of the chemical calculations involved but also ensures that even the most complex problem can be broken down into simpler, manageable steps.

Keywords

Aluminum sulfate (Al2(SO4)3), sodium aluminate, molar mass calculation, equivalent weight, chemical analysis, solution concentration.