   Chapter 4, Problem 66PS

Chapter
Section
Textbook Problem

You can dissolve an aluminum soft drink can in an aqueous base such as potassium hydroxide.2 Al(s) + 2 KOH(aq) + 6 H2O(l) → 2 KAI(OH)4(aq) + 3 H2(g)If you place 2.05 g of aluminum in a beaker with 185 mL of 1.35 M KOH, will any aluminum remain? What mass of KAI(OH)4 is produced?

Interpretation Introduction

Interpretation:

If 2.05g of Al placed in a beaker with 185mL of 1.35MKOH, remaining aluminum should be determined and the mass of KAl(OH)4 produced in the given reaction also has to be calculated.

Concept introduction:

• Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.
• Numberofmole=GivenmassofthesubstanceMolarmass
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).

Molarity=Amountofsolute(mol)Volumeofsolution(L)

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
Explanation

Balanced chemical equation for the given reaction is,

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

The amount of Al available in the reaction can be calculated as follows,

2.05gAl×1molAl26.98gAl=0.076molAl

The amount of KOH available in the reaction is,

0.185LKOH×1.35molKOH1LKOH=0.249molKOH

Here 0.076moleofAl would react completely with 0.076×(22)=0.076molofKOH but there is more KOH present in that, so KOH is in excess and Al acts as a limiting reagent

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