Highlander Help

REVIEW ACSIII

Resources:

Have you found videos, websites, or explanations that helped you understand this chapter? Let us know and we'll add them to "Resources" part of this page for other students to use.


Answer Explanations Are Now Available!    

 

Question 1

A compound is found to consist of 28.1% sodium, 13.2% boron and 58.7% oxygen. What is its simplest formula?

Select one:
 Correct

 

Explanation

Na: 28.1g / (23 g/mol) = 1.22 mol

 

B: 13.2 g/ (10.81 g/mol) = 1.22 mol


O: 58.7g/ ( 16 g/mol) = 3.67 mol


Na:  1.22 / 1.22 = 1

B: 1.22 / 1.22 = 1

O : 3.67 / 1.22 = 3


 NaBO3

 

Question 2

Which of these compounds contains the greatest mass percentage of nitrogen?

Select one:
 Correct

 

Explanation

 

CH3NH2 = 14 / (12 +3+14+2) = 0.45

NaNH2 = 14 / (14 +35 +2) = 0.27

 Al(CN)3 =  14 *3 / (27 + 36 + 42) = 0.4

 Pb(N3)2  = 84 / (207+84) = 0.28
 

Question 3

What mass of carbon is present in 0.500 mol of phenol (C6H5OH)?

Select one:
 Correct

 

Explanation

molar mass of phenol = 94.11 g/mol


mass = molar mass x moles = 94.11 x 0.500 = 47.06 g

the mass percentage of C in phenol = 6x12/94.5 = 76.2%

mass of carbon = 47.06 x 76.2% = 36 g


 

Question 4

A 4.000 g sample of an unknown metal, M, was completely burned in excess O2 to yield 0.02225 mol of the metal oxide, M2O3. What is the metal?

Select one:
 Correct

 

Explanation

 4M + 3O2 ---> 2M2O3

from this reaction we can see that the moles of M = 2 * moles of M2O3
= 2 * 0.02225 = 0.0445 mol

mass of M = 4.000 g

molar mass of M = mass / moles = 4.000 / 0.0445 = 89 g/mol

the molar mass of Y = 89 g/mol

 

Question 5

What is the maximum mass in grams of aluminum chloride that could be obtained from 2.50 mol of barium chloride and excess aluminum sulfate? This is the balanced equation for the reaction: Al2(SO4)3 + 3BaCl2--> 3BaSO4 + 2AlCl3. BaCl2 molar mass is 208.3 g/mol and AlCl3molar mass is 133.3 g/mol.

Select one:
 Correct

 

Explanation

Al2(SO4)3 + 3BaCl2--> 3BaSO4 + 2AlCl3

from this equation we know that molar ratio of BaCl2 to AlCl3 = 3:2

2.5 mol BaCl x ( 2 mol AlCl / 3 mol BaCl2)  = 1.67 mol 

molar mass of AlCl3 = 133.3 g/mol

mass of AlCl3 = moles x molar mass = 1.67 x 133.3 = 222 g
 

Question 6

A sample of a compound of xenon and fluorine contains molecules of a single type; XeFn, where n is a whole number. If 5.0 x 1020 of these XeFn molecules have a mass of 0.172 g, what is the value of n?

Select one:
 Correct

 

Explanation

 1 mole = 6.02 x 1023  molecules 

moles of XeFn = (5 x1020 ) x (1mol/  (6.02 x 1023 molecules)= 8.3 x 10-4 mol


mass of XeFn = 0.172 g


molar mass of XeFn =  0.172 g / (0.00083 mol)  = 207.2 g/mol


atomic mass of Xe = 131.3; atomic mass of F = 19.0


207.2 g/mol - 131 g/mol = 75.7 g/mol


(75.7 g/mol) / (19g Fe) = 4

 

Question 7

1.0 x 10-5 g sample of a compound is known to contain 1.36 x 1017 molecules. This compound is

Select one:
 Correct

 

Explanation

1 mole = 6.02 x 1023


# moles of compound: 1.36 x 1017 / (6.02 x 1023) = 2.26 x 10-7 mols 

molar mass compound:  1.0 x 10-5 / (2.26 x 10-7) = 44.2 g/mol 


CO2 = 44.2 g/mol

 

Question 8

1.438 g sample of a compound of nitrogen and oxygen contains 1.000 g of oxygen. What is the empirical formula?

Select one:
 Correct

 

Explanation

 N : 0.438g / 14.01 g/mol = 0.0313  --- > 0.313 /.0313 = 1


O : 1.00g / 16.00 g/mol = .0625 ---> .0625/ .0313 = 2


NO2



 

Question 9

The limiting reagent in a particular reaction can be recognized because it is the reagent that

Select one:
 Correct

 

Explanation

The limiting reagent produces the smallest quantity of the product. Therefore, it is the one that is used up first.  

 

Question 10

Consider this reaction used for the production of lead. What is the maximum mass of lead that can be obtained by the reaction of 58.0 g PbO and 30.0 g ofPbS?

Select one:
 Correct

 

Explanation

 

 

Question 11

What volume of 0.20 M BaCl2 is required to react completely with 25.0 mL of 0.600 M Na2SO4? This is the net ionic equation for the reaction:

Select one:
 Correct

 

Explanation

 MBaCl2VBaCl2 = MNa2SO4VNa2SO4


VBaCl2 = MNa2SO4VNa2SO4/MBaCl2 = (25.0 x 0.600) / 0.20 = 75 ml


75 ml

 

Question 12

Select one:
 Correct

 

Explanation

 PV = nRT 


n = PV/ RT 


n = (1* 5L) / ( .08206 * 273) = 0.223 mol CaCO3


0.223 mol CaCO3 x ( 1 mol CO2/ 1 mol CaO ) = 0.223 mol


 0.223 mol x 100 g/ mol = 22 g 

Comments