Start with 1g Aluminum and the final weight(of crystal) equals 16.845g 1.Write t

Question

Start with 1g Aluminum and the final weight(of crystal) equals 16.845g

1.Write the balanced net ionic equations for the following:(a)aluminum and potassium hydroxide, yielding [Al(OH)4]- and hydrogen gas;(b)hydrogen ions and [Al(OH)4]-, yielding aluminum hydroxide;(c)aluminum hydroxide and hydrogen ions, yielding [Al(H2O)6]3+; and (d) the formation of alum from potassium ions, sulfate ions, [Al(H2O)6]3+ and water

2.Determine the theoretical yield of the alum. use the aluminum foil as the limiting reagent and presume that the foil was pure aluminum.

3.Calculate the percent yield of your alum crystals.

Explanation / Answer

(a)
2Al + 2OH- + 6H2O ---> 2[Al(OH)4]- + 3H2

(b) H+ + [Al(OH)4]- ---> H2O + Al(OH)3

(c) Al(OH)3 + 3H2O + 3H+ ---> [Al(H2O)6]3+

(d) [Al(H2O)6]3+ + K2[SO4] ---> KAl(SO4)2 + 6H2O

since from the balanced chemcal reactions

1 mole of Al gives q mole of KAl(SO4)2

27 gram og Al gives 474.3 g of KAl(SO4)2

hence 16.845 g of Al shoud give ----

474.3X 16.845 / 27

= 295.9 g of KAl(SO4)2'

3) for % yield you need to give practical yield

as% = (pracical yield /theoriticalyield) X 100

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