The instructions for the potassium aluminum sulfate preparation give a target of
ID: 897295 • Letter: T
Question
The instructions for the potassium aluminum sulfate preparation give a target of between 0.025 and 0.05 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. Report this mass of aluminum to two significant figures. Do not include units in the answer box. 2 Al(s) + 2KOH(aq) + 6H2O(l) à 2KAl(OH)4(aq) + 3H2(g) KAl(OH)4(aq) + 2H2SO4(aq) à KAl(SO4)2(aq) + 4H2O(l)
Explanation / Answer
2Al(s) + 2KOH(aq) + 6H2O(l) ---> 2KAl(OH)4(aq) + 3H2(g)
(KAl(OH)4(aq) + 2H2SO4(aq) ---> KAl(SO4)2(aq) + 4H2O(l))*2
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Al(s) + 2KOH(aq) + 4H2SO4(aq) ----> 2KAl(SO4)2(aq) + 2 H2O(l) + 3H2(g)
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from equation , 2 mole Al = 2 mole KAl(SO4)2
No of moles of KAl(SO4)2 = 0.05 mole
No of moles of Al required = 1*0.05 = 0.05 mole
mass of Al required = 0.05*27 = 1.35 grams
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