Question

# The experiment calls for preparing one of two salts, either copper(ii) sulfate or potassium aluminum sulfate....

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) + 3H2(g)

KAl(OH)4(aq) + 2H2SO4(aq) ----> KAl(SO4)2(aq) + 4H2O(l)

eq1. 2 Al(s) + 2KOH(aq) + 6H2O(l) ---> 2KAl(OH)4(aq) + 3H2(g)

eq2.      KAl(OH)4(aq) + 2H2SO4(aq) ----> KAl(SO4)2(aq) + 4H2O(l)

If we do, eq2x2

eq3     2KAl(OH)4(aq) + 4H2SO4(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

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