Question

how many grams of aluminium were used to prepare 12.0 g of potassium alum if the percent yield was 80.0%?

Answer 1

Given

12 g potassium alum

Percent yield = 80%

Percent yield = ( actual yield / theoretical yield ) *100

Here actual yield is 12 g

So

Theoretical yield = (actual yield / percent yield) *100

= (12 g/80) * 100

= 15 g of potassium alum

Moles of potassium alum = mass of potassium alum / Molar mass of potassium alum

= 15 g/258.21 g/mol

= 0.058 moles potassium alum

Chemical reaction is

2 Al (s) + 2 KOH (aq) + 22 H2O (l) + 4 H2SO4 (aq) → 2 KAl(SO4)2 + 12 H2O (s) + 3 H2 (g)

So

2 moles aluminium produces 2 moles potassium alum

So

0.058 moles potassium alum is produced by

0.058 moles potassium alum * 2 moles aluminium/2 moles potassium alum

= 0.058 moles aluminium

So

Mass of aluminium = moles of aluminium * molar mass of aluminium

= 0.058 moles * 27 g/mol

= 1.57 g aluminium is used