Question

The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate. The instructions for the potassium aluminum sulfate preparation give a target of between 0.025 and 0.050 moles of the product. The instructions also specify to use an excess of potassium hydroxide but not more than 2 times the amount required by the balanced equation. Calculate the mass in grams of aluminum needed to prepare the maximum number of moles of product as noted above.

2 Al_(s) + 2KOH_(aq) + 6H₂O(l) ---> 2KAl(OH)_4(aq) + 3H_2(g)

KAl(OH)_4(aq) + 2H₂SO_{4(aq) ---->} KAl(SO₄)_2(aq) + 4H₂O_(l)

Answer 1

eq1. 2 Al_(s) + 2KOH_(aq) + 6H₂O(l) ---> 2KAl(OH)_4(aq) + 3H_2(g)

eq2.      KAl(OH)_4(aq) + 2H₂SO_{4(aq) ---->} KAl(SO4)2_(aq) + 4H₂O_(l)

If we do, eq2x2

eq3     2KAl(OH)_4(aq) + 4H₂SO_{4(aq) ---->} 2KAl(SO4)2_(aq) + 8H₂O_(l)

if we do, eq1 + eq3

2Al + 2KOH+ 4H2SO4 = 2KAl(SO4)2_(aq) + 2H₂O_(l) + 3H2(g)

Molar mass of KAl(SO4)2 = 258.21 g/mol

Atomic weight of Al = 27

Maximum moles of potassium aluminum sulfate can be prepared is 0.050 moles.

Again 1 mole of KAl(SO4)2 produced from 1 mol of aluminium (Al).

so, 0.05 moles will be obtained from 0.05 moles of Al.

In grams, Al required is = 0.05*27 = 1.35 g