Glencoe Chemistry: Matter and Change, Student Edition
Glencoe Chemistry: Matter and Change, Student Edition
1st Edition
ISBN: 9780076774609
Author: McGraw-Hill Education
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 11, Problem 93A

(a)

Interpretation Introduction

Interpretation:

Limiting reactant is to be calculated if 40 g of SiO2 and 40 g of HF are reacted.

Concept introduction:

1 mol is defined as 6.02214076× 1023 particles of any substance.

Molecular mass is defined as the mass of one mole of substance.

The formula to calculate moles of substance

Moles=MassofsubstanceMolarmassofsubstance

(a)

Expert Solution
Check Mark

Answer to Problem 93A

HFact as the limiting reagent.

Explanation of Solution

The given compound is SiO2.

The formula to calculate formula mass is,

Formulamass=(N1×x1)+(N2×x2)+...

Where,

  • N1 represents the number of atoms of first element in chemical formula.
  • x1 represents the atomic mass of first element.
  • N2 represents the number of atoms of second element in chemical formula.
  • x1 represents the atomic mass of second element.

The atomic mass of silicon is 28.086 g / mol.

The atomic mass of oxygen is 15.99g/mol.

Substitute the value of atomic mass in the above formula.

Formulamass=(N1×x1)+(N2×x2)+...=(1×28.086)+(2×15.99)=60.066g/mol

Molecular mass of SiO2 is 60.066 g/mol.

The given compound is HF.

The formula to calculate formula mass is,

Formulamass=(N1×x1)+(N2×x2)+...

Where,

  • N1 represents the number of atoms of first element in chemical formula.
  • x1 represents the atomic mass of first element.
  • N2 represents the number of atoms of second element in chemical formula.
  • x1 represents the atomic mass of second element.

The atomic mass of fluorine is 18.99 g / mol.

The atomic mass of hydrogen is 1.00g/mol.

Substitute the value of atomic mass in the above formula.

Formulamass=(N1×x1)+(N2×x2)+...=(1×1)+(1×18.99)=19.99g/mol

Molecular mass of HF is 19.99 g/mol.

numberofmolesofHF=MassofHFMolarmassofHF=40g19.99g/mol=2moles

numberofmolesofSiO2=MassofSiO2MolarmassofSiO2=40g60.066g/mol=0.666moles

SiO2 react with HF

SiO2(s)+6HF(aq)H2SiF6(aq)+2H2O(l)

From the stoichiometry of reaction,

For SiO2,

1molesofSiO2produce=1molesofH2SiF60.666molesofSiO2produce=1×0.666molesofH2SiF60.666molesofSiO2produce=0.666molesofH2SiF6

For HF

6molesofHFproduce=1molesofH2SiF61molesofHFproduce=16molesofH2SiF62×1molesofHFproduce=16molesofH2SiF62molesofHFproduce=0.333molesofH2SiF6

The reactant which produce least amount of product is limiting reagent. So, HFact as the limiting reagent.

(b)

Interpretation Introduction

Interpretation:

The mass of excess reactant is to be calculated

Concept introduction:

The reactant which left over after the reaction is excess reactant.

(b)

Expert Solution
Check Mark

Answer to Problem 93A

20 g of SiO2 left after completion of reaction.

Explanation of Solution

Reaction will produce H2SiF6 according to limiting reagent

Moles of H2SiF6was calculated above.

Number of moles of H2SiF6produced by limiting reagent is the actual amount of H2SiF6

formed and it is equal to 0.333 moles.

0.333 moles of H2SiF6 was produced from 0.333 moles of SiO2 So, remaining SiO2 = moles of SiO2 present − moles of SiO2 consumed

= 0.666 − 0.333 = 0.333

Massofsubstance=moles×molarmassofsubstance=0.333×60.066=20g

c)

Interpretation Introduction

Interpretation : The theoretical yield needs to be calculated.

Concept introduction : Theoretical yield is the maximum amount of product formed.

c)

Expert Solution
Check Mark

Answer to Problem 93A

Theoretical yield is 47.98 g of H2SiF6

Explanation of Solution

Step 1: Write balance chemical equation

SiO2(s)+6HF(aq)H2SiF6(aq)+2H2O(l)

Step 2: convert mass of reactant to moles

Molecular mass of HF is 19.99 g/mol.

numberofmolesofHF=MassofHFMolarmassofHF=40g19.99g/mol=2moles

Step 3: Convert moles of reactant to moles of product using mole ratio

6molesofHFproduce=1molesofH2SiF61molesofHFproduce=16molesofH2SiF62×1molesofHFproduce=2×16molesofH2SiF62molesofHFproduce=0.333molesofH2SiF6

Step 4: Convert moles of product to mass (theoretical yield)

MolarmassofH2SiF6=144.092g/moleMass=Molarmass×MolesMassofH2SiF6(Theoreticalyield)=144.092×0.333=47.98g

d)

Interpretation Introduction

Interpretation:

If 45.8g of H2SiF6 was produce, the percent yield needs to be calculated.

Concept introduction:

The percent yield can be calculated as follows:

PercentYield=ActualyieldTheoreticalyield×100

d)

Expert Solution
Check Mark

Answer to Problem 93A

Percent yield = 95.45%

Explanation of Solution

Actual yield = 45.8 g (given)

PercentYield=ActualyieldTheoreticalyield×100=45.847.98×100=95.45%

Chapter 11 Solutions

Glencoe Chemistry: Matter and Change, Student Edition

Ch. 11.2 - Prob. 11PPCh. 11.2 - Prob. 12PPCh. 11.2 - Prob. 13PPCh. 11.2 - Prob. 14PPCh. 11.2 - Prob. 15PPCh. 11.2 - Prob. 16PPCh. 11.2 - Prob. 17SSCCh. 11.2 - Prob. 18SSCCh. 11.2 - Prob. 19SSCCh. 11.2 - Prob. 20SSCCh. 11.2 - Prob. 21SSCCh. 11.2 - Prob. 22SSCCh. 11.3 - Prob. 23PPCh. 11.3 - Prob. 24PPCh. 11.3 - Prob. 25SSCCh. 11.3 - Prob. 26SSCCh. 11.3 - Prob. 27SSCCh. 11.4 - Prob. 28PPCh. 11.4 - Prob. 29PPCh. 11.4 - Prob. 30PPCh. 11.4 - Prob. 31SSCCh. 11.4 - Prob. 32SSCCh. 11.4 - Prob. 33SSCCh. 11.4 - Prob. 34SSCCh. 11.4 - Prob. 35SSCCh. 11 - Prob. 36ACh. 11 - Prob. 37ACh. 11 - Prob. 38ACh. 11 - Prob. 39ACh. 11 - Prob. 40ACh. 11 - Prob. 41ACh. 11 - Prob. 42ACh. 11 - Prob. 43ACh. 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. 50ACh. 11 - Prob. 51ACh. 11 - Prob. 52ACh. 11 - Antacids Magnesium hydroxide is an ingredient in...Ch. 11 - Prob. 54ACh. 11 - Prob. 55ACh. 11 - Prob. 56ACh. 11 - Prob. 57ACh. 11 - Prob. 58ACh. 11 - Prob. 59ACh. 11 - Ethanol (C2H5OH) , also known as grain alcohol,...Ch. 11 - Welding If 5.50 mol of calcium carbide (CaC2)...Ch. 11 - Prob. 62ACh. 11 - Prob. 63ACh. 11 - Prob. 64ACh. 11 - Prob. 65ACh. 11 - Prob. 66ACh. 11 - Prob. 67ACh. 11 - Prob. 68ACh. 11 - Prob. 69ACh. 11 - Prob. 70ACh. 11 - Prob. 71ACh. 11 - Prob. 72ACh. 11 - Prob. 73ACh. 11 - Prob. 74ACh. 11 - Prob. 75ACh. 11 - Prob. 76ACh. 11 - Prob. 77ACh. 11 - Prob. 78ACh. 11 - Prob. 79ACh. 11 - Prob. 80ACh. 11 - Prob. 81ACh. 11 - Prob. 82ACh. 11 - Prob. 83ACh. 11 - Prob. 84ACh. 11 - Prob. 85ACh. 11 - Prob. 86ACh. 11 - Prob. 87ACh. 11 - Prob. 88ACh. 11 - Prob. 89ACh. 11 - Prob. 90ACh. 11 - Lead(ll) oxide is obtained by roasting galena,...Ch. 11 - Prob. 92ACh. 11 - Prob. 93ACh. 11 - Prob. 94ACh. 11 - Prob. 95ACh. 11 - Prob. 96ACh. 11 - Prob. 97ACh. 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. 110ACh. 11 - Prob. 111ACh. 11 - Prob. 112ACh. 11 - Prob. 113ACh. 11 - Prob. 114ACh. 11 - Prob. 115ACh. 11 - Prob. 116ACh. 11 - Prob. 117ACh. 11 - Prob. 118ACh. 11 - Prob. 119ACh. 11 - Prob. 120ACh. 11 - Prob. 1STPCh. 11 - Prob. 2STPCh. 11 - Prob. 3STPCh. 11 - Prob. 4STPCh. 11 - Prob. 5STPCh. 11 - Prob. 6STPCh. 11 - Prob. 7STPCh. 11 - Prob. 8STPCh. 11 - Prob. 9STPCh. 11 - Prob. 10STPCh. 11 - Prob. 11STPCh. 11 - Prob. 12STPCh. 11 - Prob. 13STPCh. 11 - Prob. 14STPCh. 11 - Prob. 15STPCh. 11 - Prob. 16STPCh. 11 - Prob. 17STP
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