MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 6, Problem 6.97P

(a)

Interpretation Introduction

Interpretation:

The mass of AgI formed when AgNO3 reacted with NaI is to be calculated.

Concept introduction:

A limiting reagent is the species that is completely consumed in a chemical reaction. The amount of product formed or reactant reacted in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 6.97P

The mass of AgI formed is 0.35g.

Explanation of Solution

The equation of the chemical reaction between AgNO3 and NaI is,

AgNO3(aq)+NaI(aq)AgI(s)+NaNO3(aq)

The formula to calculate moles of AgNO3 is,

Moles of AgNO3=(volume of AgNO3)( mass of AgNO3molar mass of AgNO3) (1)

Substitute 50.0mL for volume of AgNO3, 5.0g/L for mass of AgNO3 and 169.9g/mol for molar mass of AgNO3 in the equation(1).

Moles of AgNO3=(50.0mL)( 5.0g/L169.9g/mol)=1.47145×103mol

The formula to calculate moles of NaI is,

Moles of NaI=(volume of NaI)( mass of NaImolar mass of NaI) (2)

Substitute 50.0mL for volume of NaI, 5.0g/L for mass of NaI and 149.9g/mol for molar mass of NaI in the equation(2).

Moles of NaI=(50.0mL)( 5.0g/L149.9g/mol)=1.677785×103mol

Moles of AgNO3 is less so AgNO3 is the imiting reagent.

The formula to calculate mass of AgI is,

Mass of AgI=(moles of AgNO3)(1molAgI1molAgNO3)(molar mass of AgI) (3)

Substitute 1.47145×103mol for moles of AgNO3 and 234.8g/mol for molar mass of AgI in the equation (3).

Mass of AgI=(1.47145×103mol)(1molAgI1molAgNO3)(234.8g/mol)=0.345496g0.35g

Conclusion

The mass of AgI formed is 0.35g.

(b)

Interpretation Introduction

Interpretation:

ΔHrxn° of the following reaction is to be determined.

Ag+(aq)+I(aq)AgI(s)

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction (ΔHrxn°) is as follows:

ΔHrxn°=mΔHf (products)°mΔHf (reactants)°

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

(b)

Expert Solution
Check Mark

Answer to Problem 6.97P

ΔHrxn° of the following reaction is 112.3kJ.

Explanation of Solution

The formula to calculate the standard enthalpy of a given reaction (ΔHrxn°) is as follows:

ΔHrxn°=[{1ΔHf°[AgI(s)]}{1ΔHf°[Ag+(aq)]+1ΔHf°[I(aq)]}] (4)

Substitute 62.38kJ/mol for ΔHf°[AgI(s)], 105.9kJ/mol for ΔHf°[Ag+(aq)] and 55.94kJ/mol for ΔHf°[I(aq)] in the equation (4).

ΔHrxn°=[{(1mol)(62.38kJ/mol)}{(1mol)(105.9kJ/mol)+(1mol)(55.94kJ/mol)}]=112.34kJ112.3kJ

Conclusion

ΔHrxn° of the following reaction is 112.3kJ.

(c)

Interpretation Introduction

Interpretation:

ΔTsol of AgI is to determined

Concept introduction:

Heat capacity (C) of a material is the quantity to measure the heat needed to raise the temperature of a substance by 1K. The formula to calculate heat required is as follows:

q=(C)(ΔT) (5)

Here,

ΔT is the temperature difference.

q is the heat released or absorbed.

C is the heat capacity of the substance.

Specific heat capacity (c) of a substance is the quantity to measure the heat needed to raise the temperature of 1g of a substance by 1K. The formula to calculate heat required is as follows:

q=(mass)(c)(ΔT) (6)

Here,

ΔT is the temperature difference.

q is the heat released or absorbed.

c is the specific heat capacity of the substance.

(c)

Expert Solution
Check Mark

Answer to Problem 6.97P

ΔTsol is 0.40K.

Explanation of Solution

The equation of the chemical reaction between AgNO3 and NaI is,

AgNO3(aq)+NaI(aq)AgI(s)+NaNO3(aq)

The formula to calculate ΔHrxn° is,

ΔHrxn°=(ΔHrxn°of AgI)(1molAgI1molAgNO3)(moles of AgNO3) (7)

Substitute 112.34kJ for ΔHrxn° of AgI and 1.47145×103mol for moles of AgNO3 in the equation (7).

ΔHrxn°=(112.34kJ)(1molAgI1molAgNO3)(1.47145×103mol)=165.3026×103kJ

The formula to calculate mass of solution is,

Mass of solution=(density of solution)(volume of AgNO3+volumeofNaI) (8)

Substitute 1g/mL for density of solution , 50mL for volume of AgNO3 and 50mL for volume of NaI in the equation (8).

Mass of solution=(1g/mL)(50mL+50mL)=100g

Rearrange the equation (6) to calculate ΔT.

ΔT=q(mass)(c) (9)

Substitute 165.3026×103kJ for q, 4.184J/gK for c and 100g for mass in the equation (9).

ΔT=165.3026×103kJ(100g)(4.184J/gK)(1000J1kJ)=0.39508K0.40K

Conclusion

ΔTsol is 0.40K.

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Chapter 6 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 6.3 - When 25.0 mL of 2.00 M HNO3 and 50.0 mL of 1.00 M...Ch. 6.3 - Prob. 6.6BFPCh. 6.3 - Prob. 6.7AFPCh. 6.3 - Prob. 6.7BFPCh. 6.4 - Prob. 6.8AFPCh. 6.4 - Prob. 6.8BFPCh. 6.5 - Prob. 6.9AFPCh. 6.5 - Prob. 6.9BFPCh. 6.6 - Prob. 6.10AFPCh. 6.6 - Prob. 6.10BFPCh. 6.6 - Prob. 6.11AFPCh. 6.6 - Prob. 6.11BFPCh. 6.6 - Prob. B6.1PCh. 6.6 - Prob. B6.2PCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - A system releases 255 cal of heat to the...Ch. 6 - What is the change in internal energy (in J) of a...Ch. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Thermal decomposition of 5.0 metric tons of...Ch. 6 - Prob. 6.15PCh. 6 - The external pressure on a gas sample is 2660...Ch. 6 - The nutritional calorie (Calorie) is equivalent to...Ch. 6 - If an athlete expends 1950 kJ/h, how long does it...Ch. 6 - Prob. 6.19PCh. 6 - Hot packs used by skiers produce heat via the...Ch. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - For each process, state whether ΔH is less than...Ch. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - What data do you need to determine the specific...Ch. 6 - Is the specific heat capacity of a substance an...Ch. 6 - Prob. 6.35PCh. 6 - Both a coffee-cup calorimeter and a bomb...Ch. 6 - Find q when 22.0 g of water is heated from 25.0°C...Ch. 6 - Calculate q when 0.10 g of ice is cooled from...Ch. 6 - A 295-g aluminum engine part at an initial...Ch. 6 - Prob. 6.40PCh. 6 - Two iron bolts of equal mass—one at 100.°C, the...Ch. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - A 30.5-g sample of an alloy at 93.0°C is placed...Ch. 6 - When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL...Ch. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - A chemist places 1.750 g of ethanol, C2H6O, in a...Ch. 6 - High-purity benzoic acid (C6H5COOH; ΔH for...Ch. 6 - Two aircraft rivets, one iron and the other...Ch. 6 - A chemical engineer burned 1.520 g of a...Ch. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Consider the following balanced thermochemical...Ch. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - When 1 mol of KBr(s) decomposes to its elements,...Ch. 6 - Prob. 6.61PCh. 6 - Compounds of boron and hydrogen are remarkable for...Ch. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Write the balanced overall equation (equation 3)...Ch. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Calculatefor each of the following: SiO2(s) +...Ch. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - The common lead-acid car battery produces a large...Ch. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.89PCh. 6 - Prob. 6.90PCh. 6 - Prob. 6.91PCh. 6 - Prob. 6.92PCh. 6 - The following scenes represent a gaseous reaction...Ch. 6 - Prob. 6.94PCh. 6 - Prob. 6.95PCh. 6 - Prob. 6.96PCh. 6 - Prob. 6.97PCh. 6 - Prob. 6.98PCh. 6 - Prob. 6.99PCh. 6 - Prob. 6.100PCh. 6 - Prob. 6.101PCh. 6 - Prob. 6.102PCh. 6 - Prob. 6.103PCh. 6 - Prob. 6.104PCh. 6 - Prob. 6.105PCh. 6 - Prob. 6.106PCh. 6 - Liquid methanol (CH3OH) canbe used as an...Ch. 6 - Prob. 6.108P
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