Solved: Limiting Reactants (Section)Aluminum hydroxide
Chapter , Problem 78E(choose chapter or problem)
Limiting Reactants (Section 3.7)
Aluminum hydroxide reacts with sulfuric acid as follows:
\(2\ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\ \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6\ \mathrm{H}_{2} \mathrm{O}(l)\)
Which is the limiting reactant when 0.500 mol \(\mathrm{Al}(\mathrm{OH})_{3}\) and 0.500 mol \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm {Al_2(SO_4)3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?
Equation Transcription:
Text Transcription:
2 Al(OH)_{3}(s)+3 H_{2}SO_{4}(aq){rightarrow}Al_{2}(SO_{4})_{3}(aq)+6 H_{2}O(l)
Al(OH)_3
H_{2}SO_4
Al_{2}(SO_{4})_3
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