Answer: The theoretical yield of alum is 12.09 g
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
We are given:
Molarity of sulfuric acid = 9.9 M
Volume of solution = 8.3 mL = 0.0083 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]\text{Moles of sulfuric acid}=(9.9mol/L\times 0.0083L)=0.08217mol[/tex]
The given chemical equation follows:
[tex]2Al+2KOH+4H_2SO_4+22H_2O\rightarrow 3H_2+2KAl(SO_4)_2.12H_2O[/tex]
By the stoichiometry of the reaction:
4 moles of sulfuric acid produces 2 moles of alum
So, 0.08217 moles of sulfuric acid will produce = [tex]\frac{2}{4}\times 0.08217=0.0411mol[/tex] of alum
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Molar mass of alum = 294.24 g/mol
Plugging values in equation 2:
[tex]\text{Mass of alum}=(0.0411mol\times 294.24g/mol)=12.09g[/tex]
Hence, the theoretical yield of alum is 12.09 g
