Chemistry 12.3 Section Review Answers

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Study Guide for Content Mastery Answer Key Chemistry: Matter and Change T193

Proper noun Engagement Class

72 Chemistry: Matter and Alter • Chapter 12 Written report Guide for Content Mastery

Section 12.4 Percent Yield

In your textbook, read nearly the yields of products.

Study the diagram and the example problem.

Example Problem: The following chemic equation represents the production of gallium

oxide, a substance used in the manufacturing of some semiconductor devices.

4Ga(due south) 1 3Otwo (thou) 0 2GatwoOiii(s)

In one experiment, the reaction yielded vii.42 g of the oxide from a 7.00-g sample of gallium.

Determine the percent yield of this reaction. The molar masses of Ga and Ga2 Othree are

69.72 thousand/mol and 187.44 yard/mol, respectively.

Use the information in the diagram and example trouble to evaluate each value or

expression below. If the value or expression is correct, write correct. If it is incorrect,

write the correct value or expression.

i. actual yield: unknown

2. mass of reactant: 7.00 g Ga

3. number of moles of reactant: 7.00 g Ga 3

four. number of moles of product: 0.100 mol Ga 3

5. theoretical yield: 0.0500 mol Gaii O3 3

vi. percent yield: three100 seven.42 k Gaii Othree / 9.37 g Ga2 Oiii 3100

9.37 g Ga2 O iii

}}

seven.42 g Gatwo O 3

right

187.44 g Ga2 O iii

}}

1 mol Ga2 O 3

2 mol Ga2 O iii

}}

1 mol Ga

7.00 one thousand Ga 31 mol Ga/69.72 g Ga

69.72 grand Ga

}}

1 mol Ga

correct

7.42 g Ga2 O 3

STUDY GUIDE FOR CONTENT MASTERY

CHAPTER 12

percent yield five 3 100%

actual yield

}}

theoretical yield

mass of product from experimental

measurement

mass of product predicted from stoichiometric adding using

a. mass of reactant

b. 4-step mass-to-mass conversion

1. Write the balanced chemic equation.

two. Calculate the number of moles of reactant, using

molar mass.

3. Calculate the number of moles of production, using

the appropriate mole ratio.

4. Calculate the mass of product, using the reciprocal

of molar mass.

0.100 mol Ga 3

2 mol Ga2 O3 / 4 mol Ga

Name Date Form

Report Guide for Content Mastery Chemistry: Matter and Change • Chapter 12 71

STUDY GUIDE FOR CONTENT MASouthTERY

CHAPTER 1two

Department 12.3 Limiting Reactants

In your textbook, read about why reactions terminate and how to make up one's mind the limiting

reactant.

Study the diagram showing a chemical reaction and the chemic equation that repre-

sents the reaction. Then complete the table. Show your calculations for questions 25–27

in the space below the table.

O2 1 2NO 0 2NO2

The molar masses of O2, NO, and NO2are 32.00 yard/mol, 30.01 thousand/mol, and 46.01 chiliad/mol,

respectively.

balanced equation mole ratio 5two mol NO/1 mol Otwo

10.00 g O2 iii1 mol O2 / 32.00 g Oii five0.3125 mol Oii

xx.00 one thousand NO 31 mol NO/30.01 grand NO v0.6664 mol NO

bodily mole ratio 50.6664 mol NO/0.3125 mol O2 5ii.132 mol NO/1.000 mol Otwo

Because the actual mole ratio of NO:Oiiis larger than the counterbalanced equation mole

ratio of NO:Oii, thereastward is an backlog of NO; O2is the limiting reactant.

Mass of NO used 50.3125 mol Otwo 32 mol NO/1 mol O 2 50.6250 mol NO

0.6250 mol NO 330.01 g NO/one mol NO 518.76 g NO

Mass of NO2produced five0.6250 mol NO2 346.01 g NO2 / 1 mol NO2 528.76 m NOtwo

Backlog NO 520.00 m NO 218.76 grand NO 5i.24 1000 NO

ane 1 01

Amount and Name

Amount of OtwoAmount of NO Corporeality of NO2Limiting Reactant of Backlog Reactant

1 molecule 2 molecules 2 molecules none none

4 molecules 4 molecules iv molecules NO 2 molecules Otwo

ii molecules 8 molecules i. 4 molecules 2. O 2 3. 4 molecules NO

1.00 mol 2.00 mol four. 2.00 mol five. none half dozen. none

4.00 mol 4.00 mol vii. 4.00 mol 8. NO 9. 2.00 mol O 2

5.00 mol 7.00 mol x. 7.00 mol 11. NO 12. 1.l mol O 2

1.00 mol 4.00 mol 13. 2.00 mol xiv. O 2 15. 2.00 mol NO

0.500 mol 0.200 mol 16. 0.200 mol 17. NO 18. 0.400 mol O 2

32.00 g 60.02 g nineteen. 92.02 1000 20. none 21. none

16.00 1000 80.00 thou 22. 46.01 1000 23. O 2 24. 50.12 g NO

10.00 one thousand xx.00 yard 25. 28.76 g 26. O ii 27. ane.24 g NO

Chemistry 12.3 Section Review Answers,

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