Question

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate. (a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (b) What mass of precipitate forms when 185.5 mL of 0.533 M NaOH is added to 627 mL of a solution that contains 15.8 g of aluminum sulfate per liter?

Answer 1

Answer:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

2.571 grams of aluminum hydroxide is precipitated.

Explanation:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

b)

Moles of NaOH = n

Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)

Molarity of the solution = 0.533 M

$n=0.533 M\times 0.1855 mL=0.09887 mol$

Moles of aluminum = n

15.8 g of aluminum sulfate per liter.

Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)

Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g

Moles of aluminum sulfate = $\frac{9.9066 g}{342 g/mol}=0.02897 mol$

Moles of NaOH = 0.09887 mol

According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :

$\frac{1}{6}\times 0.09887 mol=0.01648 mol$

This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.

According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :

$\frac{2}{6}\times 0.09887 mol=0.03296 mol$ of aluminum hydroxide

Mass of 0.03296 moles of aluminum hydroxide:

0.03296 mol × 78 g/mol = 2.571 g

2.571 grams of aluminum hydroxide is precipitated.