Question

Aluminum sulfate, known as cake alum, is used in dying leather and cloth and in purifying sewage. In basic solution, it reacts with hydroxide to form a white precipitate. What mass of precipitate forms when 163.2mL of .553 M sodium hydroxide is added to 627mL of a solution containing 0.0462 M aluminum sulfate. Express answer in grams.

Answer 1

Mass of precipitate formed = 2.34468 g

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For sodium hydroxide :

Molarity = 0.553 M

Volume = 163.2 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 163.2×10⁻³ L

Thus, moles of sodium hydroxide :

`Moles=0.553 * {163.2* 10^(-3)}\ moles`

Moles of sodium hydroxide = 0.0902 moles

For aluminum sulfate :

Molarity = 0.0462 M

Volume = 627 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.0×10⁻³ L

Thus, moles of aluminum sulfate :

`Moles=0.0462 * {627* 10^(-3)}\ moles`

Moles of aluminum sulfate = 0.02897 moles

According to the given reaction:

`6NaOH_((aq))+Al_2(SO_4)_3_((aq))\rightarrow 2Al(OH)_3_((s))+3Na_2SO_4_((aq))`

6 moles of sodium hydroxide reacts with 1 mole of aluminum sulfate

So,

1 mole of sodium hydroxide reacts with 1/6 mole of aluminum sulfate

Also,

0.0902 mole of sodium hydroxide reacts with
`(1)/(6)* 0.0902` mole of aluminum sulfate

Moles of aluminum sulfate = 0.01503 mole

Available moles of aluminum sulfate = 0.02897 mole

Limiting reagent is the one which is present in small amount. Thus, sodium hydroxide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of sodium hydroxide on reaction produces 2 moles of the precipitate

So,

1 mole of sodium hydroxide on reaction produces 2/6 moles of the precipitate

Also,

0.0902 mole of sodium hydroxide on reaction produces
`(2)/(6)* 0.0902` moles of the precipitate

Moles of aluminium hydroxide formed as precipitate = 0.03006 moles

Molar mass of aluminium hydroxide = 78 g/mol

Mass of lead(II) chloride = Moles × Molar mass =
`0.03006* 78\ g` = 2.34468 g

Mass of precipitate formed = 2.34468 g