1.Calculate the moles of aluminum contined in a 1.25g piece of scrap aluminum 2.
ID: 897916 • Letter: 1
Question
1.Calculate the moles of aluminum contined in a 1.25g piece of scrap aluminum
2.Calculate the moles of alum, KAl(SO4)2.12H2O produced from a 1.25g piece of scrap aluminum. Assume 100% yeild
3.Using the answer above, calculate the mass of alum produced from a 1.25g piece of scrap metal.
4.) What is the percent yeild of a reaction if 18.3g of KAI(SO4)2.12H2O was obtained from that same 1.25g of Al metal?
The molar mass of alum is 474.390 g/mol
equation: 2Al +2KOH + 4H2SO4 +22H2O----> 2KAI(SO4)s.12H2O+3H2
Explanation / Answer
1) no.ofmoles = 1.25/27 = 0.046 moles
2)
2Al +2KOH + 4H2SO4 +22H2O----> 2KAl(SO4)2.12H2O+3H2
for 2 moles of Al ........... ................2moles of KAl(SO4)2.12H2O
(1.25/27 = 0.046 moles) of Al ..............?
= 0.046 moles of KAl(SO4)2.12H2O
3) wt of Alum = no.of mole*mol.wt
= 0.046*474.39 = 21.822 gm.
4) % of yield = (18.3/21.822)*100 = 83.86%
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