1. (6 points) Write a balanced net ionic equation for the reaction step involvin

Question

1. (6 points) Write a balanced net ionic equation for the reaction step involving the precipitation of aluminum hydroxide. Start with a balanced chemical equation for this step reaction.

2. (6 points) Write a balanced net ionic equation for the reaction step involving the aluminum hydroxide dissolving with the addition of excess sulfuric acid. Start with a balanced chemical equation for this step reaction.

3. (6 points) Calculate the mass of alum that would be produced if exactly 1.00 g of aluminum foil was used as the starting material.

4. (2 points) Calculate the percent yield if you produced 4.5 g of Alum using 1.00 g of aluminum as your starting material.

Explanation / Answer

1. Precipitation of aluminium hydroxide

molecular equation : AlCl3(aq) + 3NaOH(aq) --> Al(OH)3(s) + 3NaCl

ionic equation : Al3+(aq) + 3Cl-(aq) + 3Na+(aq) + 3OH-(aq) ---> Al(OH)3(s) + 3Na+(aq) + 3Cl-(aq_

cancel the terms Na+ and Cl- on both sides,

net-ionic equation : Al3+(aq) + 3OH-(aq) ---> Al(OH)3(s)

2. Dissolving aluminium hydroxide

molecular equation : 2Al(OH)3(s) + 3H2SO4(aq) --> Al2(SO4)3(aq) + 6H2O(l)

ionic equation : 2Al(OH)3(s) + 6H+(aq) + 3SO4^2-(aq) --> 2Al3+(aq) + 3SO4^2-(aq) + 6H2O(l)

cancel common terms on both sides,

net-ionic equation : Al(OH)3(s) + 3H+(aq) --> Al3+(aq) + 3H2O(l)

3. mass of alum produced = 1 x 474.3884/27 = 17.57 g

4. percent yield of alum = 4.5 x 100/17.57 = 25.61%

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