   Chapter 4, Problem 16PS

Chapter
Section
Textbook Problem

Aluminum chloride AlCl3, is made by treating scrap aluminum with chlorine.2 Al(s) + 3 Cl3(g) → 2 AlCl3(s)if you begin with 2.70 g of Al and 4.05 g of Cl2, (a) Which reactant is limiting? (b) What mass of AlCl3 can be produced? (c) What mass of the excess reactant remains when the reaction is completed? (d) Set up an amounts table for this problem.

(a)

Interpretation Introduction

Interpretation:

The limiting reactant of the given reaction has to be identified.

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.

Explanation

Balanced chemical equation for the given reaction is,

2Al(s)+3Cl2(g)2AlCl3(g)

To find the mass of product formed in the reaction, the amount of each reactant involved in the given reaction should be determined.

Therefore,

AmountofAl=2.70gAl×1molAl26.98g=0.10007molAlAmountofCl2=4.05gCl2×1molCl271g=0.05704molCl2

The mass of AlCl3 expected based on each reactant is,

0

(b)

Interpretation Introduction

Interpretation:

The mass of AlCl3 produced in the reaction has to be identified.

(c)

Interpretation Introduction

Interpretation:

The mass of Al remains when all the amount of Cl2 is converted to product has to be determined.

(d)

Interpretation Introduction

Interpretation:

An amount table for the given problem has to be set up.

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