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?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. You can view video lessons to learn Limiting Reagent. Or if you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.