
In the reaction $ A{l_2}{\left( {S{O_4}} \right)_3} + 6NaOH \to 2Al{\left( {OH} \right)_3} + 3N{a_2}{\left( {S{O_4}} \right)_3} $ , how many moles of $ Al{\left( {OH} \right)_3} $ can be made with $ 2.3 $ moles of $ NaOH $ and excess $ A{l_2}{\left( {S{O_4}} \right)_3} $ ?
Answer
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Hint: Given that aluminium sulphate is in excess which means the other reactant sodium hydroxide is a limited reactant. From the moles of the limiting reactant which is sodium hydroxide, the number of moles of aluminium oxide formed can be obtained. As the formation of products bases on the limited reactant only.
Complete Step By Step Answer:
Aluminium sulphate is a compound with molecular formula of $ A{l_2}{\left( {S{O_4}} \right)_3} $ reacts with a strong base like sodium hydroxide to give aluminium oxide and sodium sulphate which is a double replacement reaction.
The balanced chemical reaction is given as $ A{l_2}{\left( {S{O_4}} \right)_3} + 6NaOH \to 2Al{\left( {OH} \right)_3} + 3N{a_2}{\left( {S{O_4}} \right)_3} $
The given reaction is a balanced chemical equation. Given that one of the reactant aluminium sulphate is in excess, which means the other reactant which is sodium hydroxide is a limited reagent.
In the above reaction, $ 6 $ moles of sodium hydroxide produced $ 2 $ moles of $ Al{\left( {OH} \right)_3} $ , which means three times less than the moles of sodium hydroxide.
Thus, when $ 2.3 $ moles of sodium hydroxide are reacted, three times less than the moles of sodium hydroxide will be formed which will be $ \dfrac{{2.3}}{3} = 0.767moles $ are produced.
In the reaction $ A{l_2}{\left( {S{O_4}} \right)_3} + 6NaOH \to 2Al{\left( {OH} \right)_3} + 3N{a_2}{\left( {S{O_4}} \right)_3} $ , $ 0.767 $ moles of $ Al{\left( {OH} \right)_3} $ can be made with $ 2.3 $ moles of $ NaOH $ and excess $ A{l_2}{\left( {S{O_4}} \right)_3} $ .
Note:
In many chemical reactions, one reactant will be in excess and the other is a limited reactant. The number of moles of product produced is based upon the number of moles of limited reactant only. Ester hydrolysis is also having two reactants in which water is a limited reactant, similar to sodium hydroxide in the above reaction.
Complete Step By Step Answer:
Aluminium sulphate is a compound with molecular formula of $ A{l_2}{\left( {S{O_4}} \right)_3} $ reacts with a strong base like sodium hydroxide to give aluminium oxide and sodium sulphate which is a double replacement reaction.
The balanced chemical reaction is given as $ A{l_2}{\left( {S{O_4}} \right)_3} + 6NaOH \to 2Al{\left( {OH} \right)_3} + 3N{a_2}{\left( {S{O_4}} \right)_3} $
The given reaction is a balanced chemical equation. Given that one of the reactant aluminium sulphate is in excess, which means the other reactant which is sodium hydroxide is a limited reagent.
In the above reaction, $ 6 $ moles of sodium hydroxide produced $ 2 $ moles of $ Al{\left( {OH} \right)_3} $ , which means three times less than the moles of sodium hydroxide.
Thus, when $ 2.3 $ moles of sodium hydroxide are reacted, three times less than the moles of sodium hydroxide will be formed which will be $ \dfrac{{2.3}}{3} = 0.767moles $ are produced.
In the reaction $ A{l_2}{\left( {S{O_4}} \right)_3} + 6NaOH \to 2Al{\left( {OH} \right)_3} + 3N{a_2}{\left( {S{O_4}} \right)_3} $ , $ 0.767 $ moles of $ Al{\left( {OH} \right)_3} $ can be made with $ 2.3 $ moles of $ NaOH $ and excess $ A{l_2}{\left( {S{O_4}} \right)_3} $ .
Note:
In many chemical reactions, one reactant will be in excess and the other is a limited reactant. The number of moles of product produced is based upon the number of moles of limited reactant only. Ester hydrolysis is also having two reactants in which water is a limited reactant, similar to sodium hydroxide in the above reaction.
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