Answer to Question #146492 in General Chemistry for Gusty

Question

3. How many grams of aluminum were used to prepare 20.0 g of potassium alum if the percent
yield was 81.5%?
2 KAl(OH)4 + 4 H2SO4 → K2SO4 + Al2(SO4)3 + 8 H2O

Answer 1

The reaction to get the potassium alum from aluminium:

$2Al + 2KOH + 22 H_2O + 4H_2SO_4= 2KAl(SO_4)_2 + 12H_2O+3H_2$

The number of mols of Al equals to number of mols of $KAl(SO4)_2$ :

$n(Al) = n(KAl(SO_4)_2)$

$n(KAl(SO_4)_2) = \frac{m(KAl(SO_4)_2)}{M(KAl(SO_4)_2)}= \frac{20.0 g}{258.2 g/mol}=0.077 mol$

From this, the mass of aluminium equals to:

$m(Al) = n(Al)\cdot M(Al) = 0.077 mol \cdot 28 g/mol = 2.17 g$

By taking into account a knowledge about percentage yield, we can compute the real mass of aluminium reacted to form potassium alum:

$m(Al) = 2.17 g \cdot 0.815 = 1.77 g$