Introduction To General, Organic, And Biochemistry
Introduction To General, Organic, And Biochemistry
12th Edition
ISBN: 9781337571357
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Chapter 6, Problem 97P
Interpretation Introduction

(a)

Interpretation:

For the given reaction, the limiting reactant should be determined.

Concept Introduction:

Molarity of solution is defined as number of moles of solute in 1 L of solution.

It is mathematically represented as follows:

M=nV

Here, n is number of moles and v is volume of solution in L.

In a chemical reaction, limiting reactant is one which is totally consumed in the completion of the reaction. It limits the amount of product formed in the reaction.

Expert Solution
Check Mark

Answer to Problem 97P

Hydrogen bromide, HBr is the limiting reactant.

Explanation of Solution

The equation should be balanced first,

Ca(s)+HBr(aq)CaBr2(aq)+H2(g)

To balance the number of hydrogen and bromine give coefficient 2 to HBr thus,

Ca(s)+ 2HBr(aq)CaBr2(aq)+H2(g)

To determine the limiting reactant, calculate the number of moles of Ca and HBr present.

The molarity of HBr is 0.325 m and volume is 115 mL. Now, number of moles of HBr can be calculated as follows:

M=nV

Rearranging,

n=M×V

Putting the values,

nHBr=(0.325 M)(1 mol/L1 M)(115 mL)(103 L1 mL)=0.03737 mol

Similarly, number of moles of Ca can be calculated as follows:

n=mM

Molar mass of Ca is 40.078 g/mol thus,

nCa=1.46 g40.078 g/mol=0.0364 mol

From the balanced chemical reaction,

Ca(s)+ 2HBr(aq)CaBr2(aq)+H2(g)

1 mol Ca reacts with 2 mol of HBr thus, for 0.0364 mol of Ca, 0.07285 mol of HBr is required. But number of moles of HBr is 0.03737 mol thus, HBr is a limiting reactant.

Interpretation Introduction

(b)

Interpretation:

The volume of hydrogen gas produced should be calculated if the vapor pressure of water at 22 C with a total pressure of 754 torr.

Concept Introduction:

The ideal gas equation is as follows:

PV=nRT

Here, p is pressure, v is volume, n is number of moles, r is Universal gas constant and t is temperature.

Expert Solution
Check Mark

Answer to Problem 97P

0.4560 L.

Explanation of Solution

The balanced chemical reaction is as follows:

Ca(s)+ 2HBr(aq)CaBr2(aq)+H2(g)

Since, HBr is a limiting reactant thus, amount of hydrogen gas produced depends on the amount of HBr.

From the balanced chemical reaction, 2 mol of HBr gives 1 mol of H2

Thus, 1 mol of HBr gives 0.5 mol of H2

The number of HBr is 0.03737 mol thus, number of moles of H2 will be 0.018685 mol.

From ideal gas equation, volume can be calculated as follows:

V=nRTP

Pressure is 754 torr or 0.9921 atm and temperature is 22 OC or 295.15 K.

Putting the values,

V=(0.018695 mol)(0.082 L atm mol1 K1)(295.15 K)(0.9921 atm)=0.4560 L

Thus, volume of hydrogen gas is 0.4560 L.

Interpretation Introduction

(c)

Interpretation:

The mass of excess reactant remain after the completion of reaction should be calculated.

Concept Introduction:

The excess reactant is the one that present in excess amounts, the amount of product does not depend on the amount of the excess reactant and it remains after the completion of the reaction.

Expert Solution
Check Mark

Answer to Problem 97P

0.71018 g of Ca remains after the completion of reaction.

Explanation of Solution

The balanced chemical reaction is as follows:

Ca(s)+ 2HBr(aq)CaBr2(aq)+H2(g)

Here, HBr is limiting reactant, the number of moles of HBr is 0.03737 mol. From the reaction, 2 mol of HBr reacts with 1 mol of Ca thus, 1 mol of HBr reacts with 0.5 mol of Ca. Thus, 0.03737 mol HBr reacts with 0.018685 mol of Ca. The number of moles of Ca present is 0.0364 mol thus, number of moles of Ca remains can be calculated as follows:

nCa,remaining=0.03640.018685=0.01772 mol

Since, molar mass of Ca is 40.078 g.mol, its mass can be calculated as follows:

m=n×M

Putting the values,

m=0.01772 mol×40.078 g/mol=0.71018 g.

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Chapter 6 Solutions

Introduction To General, Organic, And Biochemistry

Ch. 6.8 - Prob. 6.11QCCh. 6.8 - Prob. 6.12QCCh. 6.8 - Problem 6-13 What is the osmolarity of a 3.3% w/v...Ch. 6.8 - Prob. 6.14QCCh. 6 - 6-15 Answer true or false. (a) A solute is the...Ch. 6 - 6-16 Answer true or false. (a) Solubility is a...Ch. 6 - 6-17 Vinegar is a homogeneous aqueous solution...Ch. 6 - 6-18 Suppose you prepare a solution by dissolving...Ch. 6 - 6-19 In each of the following, tell whether the...Ch. 6 - 6-20 Give a familiar example of solutions of each...Ch. 6 - 6-21 Are mixtures of gases true solutions or...Ch. 6 - 6-22 Answer true or false. (a) Water is a good...Ch. 6 - 6-23 We dissolved 0.32 g of aspartic acid in 115.0...Ch. 6 - Prob. 10PCh. 6 - 6-25 A small amount of solid is added to a...Ch. 6 - 6-26 On the basis of polarity and hydrogen...Ch. 6 - Prob. 13PCh. 6 - 6-28 Which pairs of liquids are likely to be...Ch. 6 - Prob. 15PCh. 6 - 6-30 Near a power plant, warm water is discharged...Ch. 6 - 6-31 If a bottle of beer is allowed to stand for...Ch. 6 - 6-32 Would you expect the solubility of ammonia...Ch. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - 6-35 Describe how we would prepare the following...Ch. 6 - Prob. 22PCh. 6 - 6-37 Calculate the w/v percentage of each of these...Ch. 6 - 6-38 Describe how we would prepare 250 mL of 0.10...Ch. 6 - 6-39 Assuming that the appropriate volumetric...Ch. 6 - 6-40 What is the molarity of each solution? (a) 47...Ch. 6 - 6-41 A teardrop with a volume of 0.5 mL contains...Ch. 6 - Prob. 28PCh. 6 - 6-43 The label on a sparkling cider says it...Ch. 6 - Prob. 30PCh. 6 - 6-45 The label on ajar of jam says it contains 13...Ch. 6 - 6-46 A particular toothpaste contains 0.17 g NaF...Ch. 6 - 6-47 A student has a bottle labeled 0.750% albumin...Ch. 6 - 6-48 How many grams of solute are present in each...Ch. 6 - 6-49 A student has a stock solution of 30.0% w/v...Ch. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - 6-53 Dioxin is considered to be poisonous in...Ch. 6 - 6-54 An industrial wastewater contains 3.60 ppb...Ch. 6 - 6-55 According to the label on a piece of cheese,...Ch. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - 6-60 Predict which of these covalent compounds is...Ch. 6 - On the basis of Tables 6.1 and 6.2 , classify the...Ch. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - 6-67 Calculate the freezing points of solutions...Ch. 6 - 6-68 If we add 175 g of ethylene glycol, C2H6O2,...Ch. 6 - Prob. 53PCh. 6 - 6-70 In winter, after a snowstorm, salt (NaCI) is...Ch. 6 - 6-71 A 4 M acetic acid (CH3COOH) solution lowers...Ch. 6 - Prob. 56PCh. 6 - 6-73 In each case, tell which side (if either)...Ch. 6 - 6-74 An osmotic semipermeable membrane that allows...Ch. 6 - 6-75 Calculate the osmolarity of each of the...Ch. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - 6-78 (Chemical Connections 6A) Oxides of nitrogen...Ch. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - 6-82 (Chemical Connections 6C) A solution contains...Ch. 6 - 6-83 (Chemical Connections 6C) The concentration...Ch. 6 - 6-84 (Chemical Connections 6D) What is the...Ch. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 76PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - 6-91 When a cucumber is put into a saline solution...Ch. 6 - Prob. 80PCh. 6 - 6-93 Two bottles of water are carbonated, with CO2...Ch. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - 6-96 We know that a 0.89% saline (NaCI) solution...Ch. 6 - Prob. 85PCh. 6 - Prob. 86PCh. 6 - 6-99 A concentrated nitric acid solution contains...Ch. 6 - 6-100 Which will have greater osmotic pressure?...Ch. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - 6-103 A swimming pool containing 20,000. L of...Ch. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - Prob. 94PCh. 6 - Prob. 95PCh. 6 - Prob. 96PCh. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - 6-111 As noted in Section 6-8C, the amount of...Ch. 6 - 6-112 List the following aqueous solutions in...Ch. 6 - 6-113 List the following aqueous solutions in...Ch. 6 - Prob. 102P
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY