   Chapter 4, Problem 30PS

Chapter
Section
Textbook Problem

The aluminum in a 0.764-g sample of an unknown material was precipitated as aluminum hydroxide, Al(OH)3, which was then converted to Al2O3 by heating strongly. If 0.127 g of Al2O3 is obtained from the 0.764-g sample, what is the mass percent of aluminum in the sample?

Interpretation Introduction

Interpretation:

The mass percent of Al in the given sample has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

Numberofmole=MassingramMolarmass

Massingramofthesubstance=Numberofmole×Molarmass

• 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).
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
Explanation

Herein to analyze a mineral sample for the quantity of Al, the sample was precipitated as aluminium hydroxide,Al(OH)3 which was then converted to Al2O3 by heating strongly.

2mol Of Al will be present in 1 mol of Al2O3.  The stoichiometry between these two are 2:1.

From the amount of Al2O3 involved in the reaction, amount of Al can be calculated. Both of these components in this reaction react in  2:1 stoichiometric ratio.

Amount of Al2O3 = 0.127g101.96g/mol=0.00124mol

Amount of  Al = 0

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