Question

Determine the limiting reagent and calculate the theoretical mass of aluminum sulfate formed and the volume of hydrogen gas at 23^oC and 752 mm Hg formed if 10.00g of aluminum is reacted with 10.00g of sulfuric acid. Be sure to show proper formulas, balanced chemical equations, solution maps and a clear and complete set-up as instructed including proper dimensional analysis as appropriate for the problem.

Answer 1

The balanced chemical reaction is

Mass of Al =10gm

Molar mass of aluminium =27gm

Moles of aluminium =10g/27g=0.3703mol

Mass of sulphuric acid taken =10gm

Molar mass of sulphric acid =98gm

Moles of sulphuric acid =10g/98g=0.102moles

So sulphuric acid is the limiting reagent

From reaction we see

3 moles of sulphuric acid produced 1mole=342gm of aluminum sulphate

So 1mole of sulphuric acid produces 342/3 gm=114gm of aluminium sulphate

So 0.102 moles of sulphuric acid produces =0.102×114gm=11.628 gm of aluminium sulphate

From reaction also we see that 3mole of sulphuric acid produceds 3moles of hydrogen gas

So 1mole of sulphuric acid produces 3/3=1mole of hydrogen gas

So 0.102 moles of sulphuric acid produces 0.102 mole of hydrogen gas

1moles of hydrogen gas oocupies 22.4L at 25°C at 760mm of Hg

So 0.102 moles of hydrogen gas occupies 22.4×0.102=2.29 L at stp

We need to find volume V₂ at T₂ 23°C=293+23=295K and 752mm of Hg pressure P₂

Using combined gas equation

P₁V₁/T₁=P₂V₂/T₂

760 mm of Hg *2.29L/298K=V₂*752mm of Hg/295K

5.84=2.549V₂

V₂=5.84/2.549=2.291L