The first reaction involves reacting aluminum with base to form a potassium alum

Question

The first reaction involves reacting aluminum with base to form a potassium aluminum hydroxide salt and hydrogen gas. The unbalanced chemical reaction is shown below, balance the reaction so that the number of each atom on the reactants is equal to the number of each atom in the products. Then, identify the type of reaction, keeping in mind that any reaction may be more than one type. Finally, record your observations during this step of the reaction. In the second reaction, sulfuric acid is added form aluminum hydroxide, potassium sulfate and water. Starting with the balanced chemical equation, write the total ionic equation by separating any aqueous salts into their ionic components. Note that there are several polyatomic anions (not

Explanation / Answer

1. Data

Al2 + KOH + H2O = KAl(OH)4 + H2

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Balanced equation:

Al2(s) + 2 KOH(aq) + 6 H2O(l) = 2 KAl(OH)4(aq) + 3 H2(g)

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Reaction Type:

Precipitation reaction, dissolution

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Observations:

For this reaction we can look the formation of a precipitate as an aquous dissolution, the soluble compund is one who has metal alkalyne ions, in this case K+

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2. Data

From products of first dissolution step, we take the most soluble compound, that is, KAl(OH)4(aq) and next step they added H2SO4 for precipitation, the balanced equation is as follows:

2 KAl(OH)4 (aq) + H2SO4 (aq) = 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2O(l)

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From here, we can obtain information to find net ionic equation:

Al(OH)4- (aq) + H+ (aq) = Al(OH)3(s) + H2O(l)

Adding more sulfuric acid in the synthesis process, precipitate of Al(OH)3 dissolves to form soluble Al3+ ions and we can obtain;

2 Al(OH)3(s) + 3 H2SO4(aq) = Al2(SO4)3(aq) + 6 H2O(l)

The net ionic equation for this final step in the synthesys of alum is:

Al(OH)3(s) + 3 H+ (aq) = Al3+(aq) + 3 H2O(l)

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