The experiment calls for preparing one of two salts, either copper(ii) sulfate o

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) + 3H2(g)
KAl(OH)4(aq) + 2H2SO4(aq) = KAl(SO4)2(aq) + 4H2O(l)

Explanation / Answer

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))

now add both

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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