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)

2 Al(s) + 2KOH(aq) + 6H2O(l) -----------------> 2KAl(OH)4(aq) + 3H2(g) -----------> 1

multiply the second equation with 2

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

2 KAl(OH)4(aq) + 4 H2SO4(aq) ---------------> 2 KAl(SO4)2(aq) + 8 H2O(l)

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

----------------------------------------------------------------------------------------------------------------

2 Al(s) + 2KOH(aq) + 4 H2SO4(aq) ----------------->  2 KAl(SO4)2(aq) + 2 H2O (l) + 3 H2

here from balanced equation :

2 mol of Al -------------------> 2 mol of KAl(SO4)2

?? mol Al -----------------> 0.050 mol of KAl(SO4)2

moles of Al = 0.050

molar mass of Al = 27 g/mol

mass of Al = 0.05 x 27 = 1.35 g

mass of aluminium = 1.35 g

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