  # 2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 10 H2O(l) ⟶ 2 KAl(SO4)2∙12 H2O(s) + 3 H2(g) alum Using the data, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant.Descriptionmassbottle mas9.150bottle mass with aluminum pieces11.177final product and bottle masss19.377What is the theoretical yield of alum and what is the percent yield of alum?

Question

2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 10 H2O(l) ⟶ 2 KAl(SO4)2∙12 H2O(s) + 3 H2(g)

alum

Using the data, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant.

 Description mass bottle mas 9.150 bottle mass with aluminum pieces 11.177 final product and bottle masss 19.377

What is the theoretical yield of alum and what is the percent yield of alum?

check_circleExpert Solution
Step 1

Chemical equation:

2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 10 H2O(l) ⟶ 2 KAl(SO4)2∙12 H2O(s) + 3 H2(g)

Step 2

The amount of Al is 11.177-9.150 = 2.027 g

1 mol of Al = 26.98

? mol is 2.027

2.027 X 1 / 26.98 = 0.075 moles.

From the equation,

2 moles of Al can form 2 moles of alum,

so, the theoretical yield is 0.075 moles.

Step 3

The actual yield = mass of final product - the mass of the bottle

= 19.377 - 9.150 = 10.227 g

The molecular weight of alum 2KAl(SO4)2.12H2O = 474.39 g/mole.

So,

The theoretical yield = the moles of alum produced X molecular weight

&...

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