Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 8, Problem 166IP

Silane, SiH4, is the silicon analogue of methane, CH4. It is prepared industrially according to the following equations:

Si ( s )   +  3HC1 ( g )  HSiCl 3 ( l )   +  H 2 ( g ) 4HSiCl 3 ( l )    SiH 4 ( g )   +  3SiCl 4 ( l )

a. If 156 mL HSiCl3 (d = 1.34 g/mL) is isolated when 15.0 L HC1 at 10.0 atm and 35°C is used, what is the percent yield of HSiCl3?

b. When 156 mL HSiCl3 is heated, what volume of SiH4 at 10.0 atm and 35°C will be obtained if the percent yield of the reaction is 93.1%?

a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The percent yield of HSiCl3 in the given reaction has to be calculated.

Concept introduction:

  • Balance equation of a reaction is written according to law of conservation of mass.
  • Number of moles of a substance, According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

  • Equation for mass of a substance from its density and volume, 

    Mass=Density×Volume

  • Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.
  • Mole ratios between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
  • Mass of a substance produced from its number of moles is,

Number of moles×Molecularmass in grams=producedmass

  • Number of moles of a substance from its given mass is,

Number of moles=GivenmassMolecularmass

  • According to ideal gas equation for finding volume of a gas,

Volume=Numberofmoles×R×Temperaturepressure

To determine: the percent yield of HSiCl3 in the given reaction.

Answer to Problem 166IP

The percent yield of HSiCl3 is 78% .

Explanation of Solution

To find: the number of moles of HCl reacted in the given reaction.

The number of moles of HCl is 5.93mol .

The volume of HCl in the given reaction is given as 15L .

The pressure is given as 10atm .

The temperature is given as 35oC=(35+273)K=308K .

The equation for finding number of moles of a substance, According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

Therefore,

The number of moles of HCl is,

nHCl=15L×10atm0.08206Latm/Kmol×308K=5.93mol

Hence,

The number of moles of HCl is 5.93mol .

To find: the number of moles HSiCl3 in the given reaction.

The number of moles of HSiCl3 in the given reaction is 1.976mol .

The balanced chemical equation for the given reaction is,

Si+3HClHSiCl3+H2

Here,

The mole ratio between HSiCl3 and HCl is 1:3 .

The number of moles of HCl is 5.93mol , and then the number of moles HSiCl3 is,

5.93molHCl×1molHSiCl33molHCl=1.976molHSiCl3

Therefore,

Theoretical calculated number of moles of HSiCl3 in the given reaction is 1.976mol .

To determine: the percent yield of HSiCl3 in the given reaction.

The percent yield of HSiCl3 is 78% .

  • The volume of obtained HSiCl3 is given as 156mL=0.156L .
  • The density of obtained HSiCl3 is given as 1.34g/L .

    Equation for mass of a substance from its density and volume,

Mass=Density×Volume

Therefore,

The mass obtained HSiCl3 is,

1.34g/L×0.156L=209g

Hence,

The actual yield of HSiCl3 in the given reaction is 209g .

  • Theoretical calculated number of moles of HSiCl3 in the given reaction is 1.976mol .

    Mass of a substance produced from its number of moles is,

Number of moles×Molecularmass in grams=producedmass

Therefore,

The mass of HSiCl3 in the given reaction is,

1.976mol×135.45g=268g

Hence,

The theoretical yield of HSiCl3 in the given reaction is 268g .

  • Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.

    The actual yield of HSiCl3 in the given reaction is 209g .

    The theoretical yield of HSiCl3 in the given reaction is 268g .

    Therefore,

    The percent yield of HSiCl3 is,

    209g268g=78%

Hence,

The percent yield of HSiCl3 is 78% .

b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The volume of SiH4 obtained in the given reaction has to be calculated.

Concept introduction:

  • Balance equation of a reaction is written according to law of conservation of mass.
  • Number of moles of a substance, According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

  • Equation for mass of a substance from its density and volume, 

    Mass=Density×Volume

  • Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield.
  • Mole ratios between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
  • Mass of a substance produced from its number of moles is,

Number of moles×Molecularmass in grams=producedmass

  • Number of moles of a substance from its given mass is,

Number of moles=GivenmassMolecularmass

  • According to ideal gas equation for finding volume of a gas,

Volume=Numberofmoles×R×Temperaturepressure

To determine: the percent yield of HSiCl3 and the volume of SiH4 obtained in the given reaction.

Answer to Problem 166IP

The volume of SiH4 obtained in the given reaction is 0.907L .

Explanation of Solution

To find: the number of moles of SiH4 in the given reaction 4HSiCl3SiH4+3SiCl4 .

The number of moles of SiH4 in the given reaction is 0.386mol .

  • The actual yield of HSiCl3 in the given reaction is 209g .

    Equation of Number of moles of a substance from its given mass is,

Number of moles=GivenmassMolecularmass

Therefore,

The number of moles of HSiCl3 in the given reaction is,

Number of moles=209g135.45g=1.543mol

  • The balanced chemical equation for the given reaction is,

    4HSiCl3SiH4+3SiCl4

    Here,

    The mole ratio between SiH4 and HSiCl3 is 1:4 .

    The number of moles HSiCl3 is calculated as 1.543mol , and then the number of moles SiH4 is,

1.543molHSiCl3×1molSiH44molHSiCl3=0.386molSiH4

Hence,

The theoretically calculated number of moles of SiH4 in the given reaction is 0.386mol .

To determine: the number of moles of SiH4 obtained in the given reaction from its given percent yield.

The number of moles of SiH4 obtained in the given reaction is 0.359mol .

The Percent yield of SiH4 in the given reaction is given as 93.1% .

Theoretically calculated number of moles of SiH4 in the given reaction is 0.386mol .

Therefore,

The actual number of moles of SiH4 in the given reaction is,

0.386mol×0.931=0.359mol

Hence,

The number of moles of SiH4 obtained in the given reaction is 0.359mol .

To determine: the volume of SiH4 obtained in the given reaction 4HSiCl3SiH4+3SiCl4 .

The volume of SiH4 obtained in the given reaction is 0.907L .

The number of moles of SiH4 obtained in the given reaction is 0.359mol .

The pressure is given as 10atm .

The temperature is given as 35oC=(35+273)K=308K .

The equation for finding volume of a gas is,

According to ideal gas equation,

Volume=Numberofmoles×R×Temperaturepressure

Therefore,

Volumeof SiH4=0.359mol×0.08206Latm/Kmol×308K10atm =0.907L

Hence,

The volume of SiH4 obtained in the given reaction is 0.907L .

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Chapter 8 Solutions

Chemistry: An Atoms First Approach

Ch. 8 - Prob. 3ALQCh. 8 - Prob. 4ALQCh. 8 - Prob. 6ALQCh. 8 - Prob. 8ALQCh. 8 - Prob. 11ALQCh. 8 - Prob. 12ALQCh. 8 - Prob. 15ALQCh. 8 - Prob. 16ALQCh. 8 - Draw molecular-level views that show the...Ch. 8 - Prob. 20QCh. 8 - Prob. 21QCh. 8 - Prob. 22QCh. 8 - Prob. 23QCh. 8 - Prob. 24QCh. 8 - Prob. 25QCh. 8 - Consider two different containers, each filled...Ch. 8 - Prob. 27QCh. 8 - Prob. 28QCh. 8 - Prob. 29QCh. 8 - Prob. 30QCh. 8 - Prob. 31QCh. 8 - Prob. 32QCh. 8 - Prob. 33QCh. 8 - Prob. 34QCh. 8 - Prob. 35QCh. 8 - Prob. 36QCh. 8 - Prob. 37ECh. 8 - Prob. 38ECh. 8 - A sealed-tube manometer (as shown below) can be...Ch. 8 - Prob. 40ECh. 8 - A diagram for an open-tube manometer is shown...Ch. 8 - Prob. 42ECh. 8 - Prob. 43ECh. 8 - Prob. 44ECh. 8 - Prob. 45ECh. 8 - Prob. 46ECh. 8 - Prob. 47ECh. 8 - Prob. 48ECh. 8 - Prob. 49ECh. 8 - Prob. 50ECh. 8 - The Steel reaction vessel of a bomb calorimeter,...Ch. 8 - A 5.0-L flask contains 0.60 g O2 at a temperature...Ch. 8 - Prob. 53ECh. 8 - A person accidentally swallows a drop of liquid...Ch. 8 - A gas sample containing 1.50 moles at 25C exerts a...Ch. 8 - Prob. 56ECh. 8 - Prob. 57ECh. 8 - What will be the effect on the volume of an ideal...Ch. 8 - Prob. 59ECh. 8 - Prob. 60ECh. 8 - An ideal gas is contained in a cylinder with a...Ch. 8 - Prob. 62ECh. 8 - A sealed balloon is filled with 1.00 L helium at...Ch. 8 - Prob. 64ECh. 8 - Consider the following reaction:...Ch. 8 - A student adds 4.00 g of dry ice (solid CO2) to an...Ch. 8 - Air bags are activated when a severe impact causes...Ch. 8 - Concentrated hydrogen peroxide solutions are...Ch. 8 - In 1897 the Swedish explorer Andre tried to reach...Ch. 8 - Sulfur trioxide, SO3, is produced in enormous...Ch. 8 - A 15.0-L rigid container was charged with 0.500...Ch. 8 - An important process for the production of...Ch. 8 - Consider the reaction between 50.0 mL liquid...Ch. 8 - Urea (H2NCONH2) is used extensively as a nitrogen...Ch. 8 - Prob. 75ECh. 8 - Prob. 76ECh. 8 - Prob. 77ECh. 8 - A compound has the empirical formula CHCl. A...Ch. 8 - Prob. 79ECh. 8 - Given that a sample of air is made up of nitrogen,...Ch. 8 - Prob. 81ECh. 8 - Prob. 82ECh. 8 - Prob. 83ECh. 8 - Prob. 84ECh. 8 - Consider the flasks in the following diagram. What...Ch. 8 - Consider the flask apparatus in Exercise 85, which...Ch. 8 - Prob. 87ECh. 8 - At 0C a 1.0-L flask contains 5.0 102 mole of N2,...Ch. 8 - Prob. 89ECh. 8 - A tank contains a mixture of 52.5 g oxygen gas and...Ch. 8 - Prob. 91ECh. 8 - Helium is collected over water at 25C and 1.00 atm...Ch. 8 - At elevated temperatures, sodium chlorate...Ch. 8 - Xenon and fluorine will react to form binary...Ch. 8 - Methanol (CH3OH) can be produced by the following...Ch. 8 - In the Mthode Champenoise, grape juice is...Ch. 8 - Hydrogen azide, HN3, decomposes on heating by the...Ch. 8 - Prob. 98ECh. 8 - Some very effective rocket fuels are composed of...Ch. 8 - The oxides of Group 2A metals (symbolized by M...Ch. 8 - Prob. 101ECh. 8 - Prob. 102ECh. 8 - Prob. 103ECh. 8 - Prob. 104ECh. 8 - Prob. 105ECh. 8 - Prob. 106ECh. 8 - Prob. 107ECh. 8 - Prob. 108ECh. 8 - Prob. 109ECh. 8 - Prob. 110ECh. 8 - Prob. 111ECh. 8 - Prob. 112ECh. 8 - Prob. 113ECh. 8 - Prob. 114ECh. 8 - Prob. 115ECh. 8 - Prob. 116ECh. 8 - Prob. 117ECh. 8 - Prob. 118ECh. 8 - Prob. 119ECh. 8 - Prob. 120ECh. 8 - Prob. 121ECh. 8 - Prob. 122ECh. 8 - Prob. 123AECh. 8 - At STP, 1.0 L Br2 reacts completely with 3.0 L F2,...Ch. 8 - Prob. 125AECh. 8 - A 2.747g sample of manganese metal is reacted with...Ch. 8 - Prob. 127AECh. 8 - Cyclopropane, a gas that when mixed with oxygen is...Ch. 8 - The nitrogen content of organic compounds can be...Ch. 8 - Prob. 130AECh. 8 - A 15.0L tank is filled with H2 to a pressure of...Ch. 8 - A spherical glass container of unknown volume...Ch. 8 - Prob. 133AECh. 8 - A 20.0L stainless steel container at 25C was...Ch. 8 - Metallic molybdenum can be produced from the...Ch. 8 - Prob. 136AECh. 8 - Prob. 137AECh. 8 - One of the chemical controversies of the...Ch. 8 - An organic compound contains C, H, N, and O....Ch. 8 - Prob. 140AECh. 8 - The total volume of hydrogen gas needed to fill...Ch. 8 - Prob. 142AECh. 8 - Prob. 143CWPCh. 8 - Prob. 144CWPCh. 8 - A certain flexible weather balloon contains helium...Ch. 8 - Prob. 146CWPCh. 8 - A 20.0L nickel container was charged with 0.859...Ch. 8 - Consider the unbalanced chemical equation below:...Ch. 8 - Prob. 149CWPCh. 8 - Which of the following statements is(are) true? a....Ch. 8 - A chemist weighed out 5.14 g of a mixture...Ch. 8 - A mixture of chromium and zinc weighing 0.362 g...Ch. 8 - Prob. 153CPCh. 8 - You have an equimolar mixture of the gases SO2 and...Ch. 8 - Methane (CH4) gas flows into a combustion chamber...Ch. 8 - Prob. 156CPCh. 8 - Prob. 157CPCh. 8 - Prob. 158CPCh. 8 - You have a helium balloon at 1.00 atm and 25C. You...Ch. 8 - Prob. 160CPCh. 8 - You are given an unknown gaseous binary compound...Ch. 8 - Prob. 162CPCh. 8 - Calculate w and E when 1 mole of a liquid is...Ch. 8 - The preparation of NO2(g) from N2(g) and O2(g) is...Ch. 8 - In the presence of nitric acid, UO2+ undergoes a...Ch. 8 - Silane, SiH4, is the silicon analogue of methane,...Ch. 8 - Prob. 167IPCh. 8 - Prob. 168IPCh. 8 - Prob. 169MP
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