(a) Answer Br 2 is limiting reactant. Explanation First convert the reactant in moles. Using relation Moles = M a s s o f C o m p o u n d M o l a r M a s s o f C o m p o u n d Molar mass of lithium is 6.93 g/mol Molar mass of Bromine gas is 159.82 g/mol Moles of lithium = 25 6.93 = 3.60 Moles of Bromine = 25 159.82 = 0.156 According to stoichiometry of reaction, 2 moles of lithium require 1 mole of bromine gas for complete reaction. And here Li is present in excess and left behind after completion of reaction. So, Br 2 is limiting reactant. (b) Interpretation Introduction Interpretation: Mass of lithium bromide produced to be determined Concept introduction: Lithium react with bromine spontaneously to produce lithium bromide. 2 L i + B r 2 → 2 L i B r Answer 27.09 g of LiBr Explanation From reaction 1 mole of Br 2 form 2 moles of LiBr. 0.156 mole of Br 2 form 2 times the mole of LiBr. = 0.312 mole of LiBr Molar mass of LiBr is 86.85g/mol Mass of LiBr formed is: m = 0.312 × 86.85 = 27.09 g o f L i B r (c) Interpretation Introduction Interpretation: Excess reactant and excess mass to be determined. Concept introduction: Lithium react with bromine spontaneously to produce lithium bromide. 2 L i + B r 2 → 2 L i B r Answer 3.288 mole of Li is in excess Mass of excess Li = 22.78 g Explanation 2 moles of Li will form 2 moles of LiBr Thus, 0.312 mole of LiBr was formed from 0.312 mole of Li. Also, 3.60 mole of lithium is present. From that amount 0.312 mole reacts, hence remaining Li = 3.60 − 0.312 = 3.288 mole 3.288 mole of Li is in excess Mass of excess Li = 3.288 × 6.93 = 22.78 g

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ISBN: 9780078746376

Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter 11, Problem 82A

(a)

Interpretation Introduction

Interpretation:

For the given reaction, limiting reactant is to be determined.

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2→2LiBr$

Lithium is a metal. Bromine is a gas and non-metal.

(a)

Expert Solution

Answer to Problem 82A

Br₂ is limiting reactant.

Explanation of Solution

First convert the reactant in moles.

Using relation

$Moles = Mass of CompoundMolar Mass of Compound$

Molar mass of lithium is 6.93 g/mol

Molar mass of Bromine gas is 159.82 g/mol

$Moles of lithium = 256.93=3.60Moles of Bromine =25159.82=0.156$

According to stoichiometry of reaction, 2 moles of lithium require 1 mole of bromine gas for complete reaction. And here Li is present in excess and left behind after completion of reaction. So, Br₂ is limiting reactant.

(b)

Interpretation Introduction

Interpretation:

Mass of lithium bromide produced to be determined

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2→2LiBr$

(b)

Expert Solution

Answer to Problem 82A

27.09 g of LiBr

Explanation of Solution

From reaction 1 mole of Br₂ form 2 moles of LiBr.

0.156 mole of Br₂ form 2 times the mole of LiBr.

= 0.312 mole of LiBr

Molar mass of LiBr is 86.85g/mol

Mass of LiBr formed is:

$m=0.312 × 86.85=27.09 g of LiBr$

(c)

Interpretation Introduction

Interpretation:

Excess reactant and excess mass to be determined.

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2→2LiBr$

(c)

Expert Solution

Answer to Problem 82A

3.288 mole of Li is in excess

Mass of excess Li = $22.78 g$

Explanation of Solution

2 moles of Li will form 2 moles of LiBr

Thus, 0.312 mole of LiBr was formed from 0.312 mole of Li.

Also, 3.60 mole of lithium is present.

From that amount 0.312 mole reacts, hence remaining Li = 3.60 − 0.312 = 3.288 mole

3.288 mole of Li is in excess

Mass of excess Li =

$3.288 ×6.93 = 22.78 g$

Chapter 11, Problem 81A

Chapter 11, Problem 83A

Chapter 11 Solutions

Chemistry: Matter and Change

Ch. 11.1 - Prob. 1PP Ch. 11.1 - Prob. 2PP Ch. 11.1 - Prob. 3PP Ch. 11.1 - Prob. 4PP Ch. 11.1 - Prob. 5SSC Ch. 11.1 - Prob. 6SSC Ch. 11.1 - Prob. 7SSC Ch. 11.1 - Prob. 8SSC Ch. 11.1 - Prob. 9SSC Ch. 11.1 - Prob. 10SSC

Ch. 11.2 - Prob. 11PP Ch. 11.2 - Prob. 12PP Ch. 11.2 - Prob. 13PP Ch. 11.2 - Prob. 14PP Ch. 11.2 - Prob. 15PP Ch. 11.2 - Prob. 16PP Ch. 11.2 - Prob. 17SSC Ch. 11.2 - Prob. 18SSC Ch. 11.2 - Prob. 19SSC Ch. 11.2 - Prob. 20SSC Ch. 11.2 - Prob. 21SSC Ch. 11.2 - Prob. 22SSC Ch. 11.3 - Prob. 23PP Ch. 11.3 - Prob. 24PP Ch. 11.3 - Prob. 25SSC Ch. 11.3 - Prob. 26SSC Ch. 11.3 - Prob. 27SSC Ch. 11.4 - Prob. 28PP Ch. 11.4 - Prob. 29PP Ch. 11.4 - Prob. 30PP Ch. 11.4 - Prob. 31SSC Ch. 11.4 - Prob. 32SSC Ch. 11.4 - Prob. 33SSC Ch. 11.4 - Prob. 34SSC Ch. 11.4 - Prob. 35SSC Ch. 11 - Prob. 36A Ch. 11 - Prob. 37A Ch. 11 - Prob. 38A Ch. 11 - Prob. 39A Ch. 11 - Prob. 40A Ch. 11 - Prob. 41A Ch. 11 - Prob. 42A Ch. 11 - Prob. 43A Ch. 11 - Interpret the following equation in terms of...Ch. 11 - Smelting When tin(IV) oxide is heated with carbon...Ch. 11 - When solid copper is added to nitric acid, copper...Ch. 11 - When hydrochloric acid solution reacts with lead...Ch. 11 - When aluminum is mixed with iron(lll) oxide, iron...Ch. 11 - Solid silicon dioxide, often called silica, reacts...Ch. 11 - Prob. 50A Ch. 11 - Prob. 51A Ch. 11 - Prob. 52A Ch. 11 - Antacids Magnesium hydroxide is an ingredient in...Ch. 11 - Prob. 54A Ch. 11 - Prob. 55A Ch. 11 - Prob. 56A Ch. 11 - Prob. 57A Ch. 11 - Prob. 58A Ch. 11 - Prob. 59A Ch. 11 - Ethanol (C2H5OH) , also known as grain alcohol,...Ch. 11 - Welding If 5.50 mol of calcium carbide (CaC2)...Ch. 11 - Prob. 62A Ch. 11 - Prob. 63A Ch. 11 - Prob. 64A Ch. 11 - Prob. 65A Ch. 11 - Prob. 66A Ch. 11 - Prob. 67A Ch. 11 - Prob. 68A Ch. 11 - Prob. 69A Ch. 11 - Prob. 70A Ch. 11 - Prob. 71A Ch. 11 - Prob. 72A Ch. 11 - Prob. 73A Ch. 11 - Prob. 74A Ch. 11 - Prob. 75A Ch. 11 - Prob. 76A Ch. 11 - Prob. 77A Ch. 11 - Prob. 78A Ch. 11 - Prob. 79A Ch. 11 - Prob. 80A Ch. 11 - Prob. 81A Ch. 11 - Prob. 82A Ch. 11 - Prob. 83A Ch. 11 - Prob. 84A Ch. 11 - Prob. 85A Ch. 11 - Prob. 86A Ch. 11 - Prob. 87A Ch. 11 - Prob. 88A Ch. 11 - Prob. 89A Ch. 11 - Prob. 90A Ch. 11 - Lead(ll) oxide is obtained by roasting galena,...Ch. 11 - Prob. 92A Ch. 11 - Prob. 93A Ch. 11 - Prob. 94A Ch. 11 - Prob. 95A Ch. 11 - Prob. 96A Ch. 11 - Prob. 97A Ch. 11 - Ammonium sulfide reacts With copper(ll) nitrate in...Ch. 11 - Fertilizer The compound calcium cyanamide (CaNCN)...Ch. 11 - When copper(ll) oxide is heated in the presence Of...Ch. 11 - Air Pollution Nitrogen monoxide, which is present...Ch. 11 - Electrolysis Determine the theoretical and percent...Ch. 11 - Iron reacts with oxygen as Shown....Ch. 11 - Analyze and Conclude In an experiment, you obtain...Ch. 11 - Observe and Infer Determine whether each reaction...Ch. 11 - Design an Experiment Design an experiment that can...Ch. 11 - Apply When a campfire begins to die down and...Ch. 11 - Apply Students conducted a lab to investigate...Ch. 11 - When 9.59 g of a certain vanadium oxide is heated...Ch. 11 - Prob. 110A Ch. 11 - Prob. 111A Ch. 11 - Prob. 112A Ch. 11 - Prob. 113A Ch. 11 - Prob. 114A Ch. 11 - Prob. 115A Ch. 11 - Prob. 116A Ch. 11 - Prob. 117A Ch. 11 - Prob. 118A Ch. 11 - Prob. 119A Ch. 11 - Prob. 120A Ch. 11 - Prob. 1STP Ch. 11 - Prob. 2STP Ch. 11 - Prob. 3STP Ch. 11 - Prob. 4STP Ch. 11 - Prob. 5STP Ch. 11 - Prob. 6STP Ch. 11 - Prob. 7STP Ch. 11 - Prob. 8STP Ch. 11 - Prob. 9STP Ch. 11 - Prob. 10STP Ch. 11 - Prob. 11STP Ch. 11 - Prob. 12STP Ch. 11 - Prob. 13STP Ch. 11 - Prob. 14STP Ch. 11 - Prob. 15STP Ch. 11 - Prob. 16STP Ch. 11 - Prob. 17STP

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For the given reaction, limiting reactant is to be determined. Concept introduction: Lithium react with bromine spontaneously to produce lithium bromide. 2 L i + B r 2 → 2 L i B r Lithium is a metal. Bromine is a gas and non-metal.

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Answer to Problem 82A

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Answer to Problem 82A

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Answer to Problem 82A

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Chapter 11 Solutions

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