Question

Sodium hydroxide interacted with 273.6g of aluminum sulfate.

Required: mass of precipitate obtained

Answer 1

Balanced equation

Al₂(SO₄)₃+NaOH$\implies$2 Al(OH)₃+3Na₂SO₄

Given mass of Sodium hydroxide as 273.6g then:

Moles of Al₂(SO₄)₃=$\frac{Mass}{MolarMass}$

=$\frac{273.6g}{342.15g/moles}$

=0.799$Moles$

Mole ratio of Al₂(SO₄)₃:2Al(OH₃)

Therefore Mole Ratio is 1:2

If 1=0.799moles

$\therefore$ 2=?

$\frac{2×0.799moles}{1}$

=$1.598moles$

But moles of the precipitae are 1.598moles

Therefore Mass=Molar mass×Moles

=$78g/mol×1.598moles$

=124.65g

Mass of precipitate obtained is therefore 124.65g