Find the ** limiting reactant**, which is the reactant that forms the less amount of product.

- determines the
**maximum amount of the product** - limiting reactant → gets used up once reaction is complete

Balanced equation**: 2 Al(OH) _{3}(s) + 3 H_{2}SO_{4}(aq) → Al_{2}(SO_{4})_{3}(aq) + 6 H_{2}O(l)**

**Mole to mole comparison:**

**1 mol****Al(OH)**_{3}forms 1 mol**Al**_{2}(SO_{4})_{3}**3 moles****H**_{2}SO_{4 }**forms 1 mol****Al**_{2}(SO_{4})_{3}

Since moles of ** **Al(OH)_{3} and H_{2}SO_{4} are **equal ****→**** ****limiting reactant = ****H _{2}SO**

**excess reactant =****Al(OH)**_{3}

The **moles of Al_{2}(SO_{4})_{3}** formed under the given conditions is:

$\mathbf{0}\mathbf{.}\mathbf{550}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{Al}}_{\mathbf{2}}{\left({\mathrm{SO}}_{4}\right)}_{\mathbf{3}}}{\mathbf{3}{\mathbf{}}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}}}\mathbf{=}$**0.183 mol Al_{2}(SO_{4})_{3}**

You may want to reference (Pages 106 - 109)Section 3.7 while completing this problem.

Aluminum hydroxide reacts with sulfuric acid as follows: 2 Al(OH)_{3}(s) + 3 H_{2}SO_{4}(aq) → Al_{2}(SO_{4})_{3}(aq) + 6 H_{2}O(l)

0.550 moles Al(OH)_{3} and 0.550 moles H_{2}SO_{4} are allowed to react. How many moles of the excess reactant remain after the completion of the reaction?

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