Question

If a student needed to make 25.678 g alum and KOH and H2SO4 are in excess,...

If a student needed to make 25.678 g alum and KOH and H2SO4 are in excess, how much Al should the students start with knowing that they would only get an 80% yield of alum?

*There was an answer to this on Chegg, but I wasn’t too sure bc they used Al2(SO4)3 as the alum chemical formula, but I thought it was KAl (SO4)2 • 12H2O)
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Answer #1

Ans :

Percent yield is 80%

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

80 = ( 25.678 g / m ) x 100

m = 32.0975 g

mol alum = mass / molar mass

= 32.0975 g / 474.39 g/mol

= 0.06766 mol

Each mol alum has one mol Al

So mol Al = 0.06766 mol

Mass Al = mol x molar mass

= 0.06766 mol x 26.98 g/mol

= 1.825 g Al

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