Unit 5b Test  Review Sheet

Short Answer

	1.	If a total of 13.5 mol of NaHCO and 4.5 mol of CHO react, how many moles of CO and NaCHO will be produced?
3NaHCO(aq) + CHO(aq)  3CO(g) + 3HO(s) +NaCHO(aq)

	2.	If 8.00 mol of NHreacted with 14.0 mol of O, how many moles of HO will be produced?
4NH(g) + 7O(g)  4NO + 6HO(g)

	3.	If 8.6 L of H reacted with 4.3 L of Oat STP, what is the volume of the gaseous water collected (assuming that none of it condenses)?
2H(g) + O(g)  2HO(g)

	4.	What is the mole ratio of D to A in the generic chemical reaction?
2A + B  C + 3D

	5.	Assuming STP and a stoichiometric amount of NH and NO in an expandable container originally at 15 L, what is the final volume if the reaction goes to completion?
4NH(g) + 6NO(g)  5N(g) + 6HO(g)

	6.	How many grams of CO are needed to react with an excess of FeO to produce 209.7 g Fe? 
FeO(s) + 3CO(g) 3CO(g) + 2Fe(s)

	7.	How many liters of O are needed to react completely with 45.0 L of HS at STP? 
2HS(g) + 3O(g)  2SO(g) + 2HO(g)

	8.	If 5.0 g of H are reacted with excess CO, how many grams of CHOH are produced, based on a yield of 86%?
CO(g) + 2H(g)  CHOH(l)

	9.	For the reaction 2Na(s) + Cl(g)  2NaCl(s), how many grams of NaCl could be produced from 103.0 g of Na and 13.0 L of Cl (at STP)?

	10.	Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas are formed when 48.7 g of sodium are added to water? 
2Na + 2HO  2NaOH + H

	11.	The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed to produce 10.0 L of O?
2KClO(s)  2KCl(s) + 3O(g)

	12.	Consider the following reaction:

2HS(g) + 3O(g)  2SO(g) + 2HO(g)

If O was the excess reagent, 8.3 mol of HS were consumed, and 137.1 g of water were collected after the reaction has gone to completion, what is the percent yield of the reaction? 
Problem

	1.	Titanium dioxide (TiO2) is an industrial chemical used as a white pigment in paint. The conversion of volatile TiCl4 to TiO2 occurs according to the reaction, . If 15.70 g of TiCl4 reacts in excess oxygen to form 10.40 g Cl2, what is the percent yield of the reaction?

	2.	Phosphorus pentachloride is formed when 17.2 g of chlorine gas react with 23.2 g of solid phosphorus (P2). Determine the reactant that is in excess.

	3.	In a reaction, 82.00 g of sodium reacts with 74.00 g of ferric oxide to form sodium oxide and iron metal. Calculate the mass of solid iron produced.

	4.	In a reaction, 10.76 g of CaCO3, 10.51 g of HCl, and excess water produced 10.26 g of CaCl26H2O. Calculate the theoretical yield of calcium chloride hexahydrate.

	5.	What is the percent yield for a reaction if the theoretical yield of C6H12 is 21 g and the actual yield recovered is only 3.8 g? 
Essay

	1.	Aluminum metal is burned in oxygen gas.
a. Write the balanced equation for this reaction.
b. If 5.433 g of aluminum is burned with 8.834 g of oxygen gas, what is the limiting reagent? 
c. What mass of product can be made?

	2.	You have 83.6 grams of H2 and 257 grams of N2 which combine according to the following equation:

N2 + 3H2 ® 2NH3

Which reactant is the limiting reagent? Explain how you can tell.

	3.	8.344 grams of sodium hydroxide is reacted with 14.290 grams of magnesium nitrate, according to the reaction NaOH + Mg(NO3)2 ® NaNO3 + Mg(OH)2.
a) Write the balanced chemical equation for this reaction.
b) Identify the limiting reagent.
c) Determine the theoretical yield of magnesium hydroxide.

	4.	In the equation for the combustion of butane, show that the law of conservation of mass is observed. Interpret the equation for the combustion of butane in terms of:
a.	Representative particles
b.	Moles
c.	Mass

	5.	A 45.00-g sample of silver nitrate is mixed with 55.00 g of hydrochloric acid to form a white precipitate of silver chloride. After the solution is filtered and dried, a white precipitate of mass 33.50 g is collected.
a.	Determine the limiting reactant.
b.	Determine the theoretical yield of silver chloride.
c.	Determine the percent yield of silver chloride.

PROBLEM SOLVING

Use the skills you have developed in this chapter to solve each problem.

	6.	When potassium chlorate, KClO3, is heated, it decomposes to produce potassium chloride and oxygen gas. How many grams of oxygen gas will be produced when 2.50 moles of potassium chlorate is decomposed? (atomic masses: O = 16.0; Cl = 35.5; K = 39.1)

	7.	Aluminum reacts with a solution of copper(II) chloride to produce copper metal and aluminum chloride, AlCl3. How many moles of copper are produced when 4.30 grams of aluminum reacts? (atomic masses: Al = 27.0; Cl = 35.5; Cu = 63.5)

	8.	When methane gas, CH4, burns completely in oxygen, the products of the reaction are carbon dioxide and water. What mass of oxygen must react in order for 4.25 grams of carbon dioxide to be produced? (atomic masses: H = 1.0; C = 12.0; O = 16.0)

	9.	Nitrogen gas reacts with hydrogen gas to produce ammonia, NH3. What mass of ammonia is produced when 6.50 grams of nitrogen reacts with hydrogen? (atomic masses: H = 1.0; N = 14.0)

	10.	Iodine reacts with concentrated nitric acid according to the following equation.
10 HNO3    +     I2     ®     2 HIO3     +    10 NO2    +    4 H2O
What mass of HIO3 is produced when 15.0 grams of iodine reacts with nitric acid? (atomic masses: H = 1.0; N = 14.0; O = 16.0; I = 126.9)

	11.	Magnesium nitride, Mg3N2, according to the following equation, reacts with water to produce magnesium hydroxide and ammonia, NH3:
Mg3N2 + 6 H2O ® 3 Mg(OH)2 + 2 NH3.
What volume of ammonia at STP will be produced when 3.00 grams of magnesium nitride reacts with water? (atomic masses: H = 1.0; N = 14.0; O = 16.0; Mg = 24.3)

	12.	Copper metal reacts with a solution of silver nitrate, AgNO3, to produce copper(II) nitrate and silver metal. In carrying out this reaction, a piece of copper wire was immersed in a solution of silver nitrate until the reaction stopped. The original mass of the wire was 3.17 grams. After the reaction stopped, the mass of the wire was 2.04 grams. What mass of silver metal was produced?

	13.	At STP, a 1.0-L sample of a gas has a mass of 2.81 grams. What is the molecular mass of this gas?

	14.	Hydrogen gas reacts with chlorine gas to form hydrogen chloride gas, HCl. What volume of hydrogen chloride at STP will be produced when 0.300 mole of chlorine reacts with hydrogen?

	15.	Iron(II) sulfide, FeS, reacts with hydrochloric acid, HCl, to produce hydrogen sulfide gas, H2S, and a solution of iron(II) chloride, FeCl2. How many moles of hydrochloric acid must react with FeS in order to produce 2.50 L of hydrogen sulfide at STP?

	16.	When propane, C3H8, reacts completely with oxygen, carbon dioxide and water are produced. How many milliliters of carbon dioxide will be produced if 200mL of oxygen reacts with propane? The volumes of both gases are at STP.

	17.	Aluminum metal reacts with dilute hydrochloric acid, HCl, to form hydrogen gas and a solution of aluminum chloride, AlCl3. What volume of hydrogen at STP will be produced when 35.0 grams of aluminum reacts with HCl?

	18.	When Pb(NO3)2 (aq) is mixed with NaCl (aq), a precipitate of PbCl2 (s) is formed. A solution containing 8.00 grams of Pb(NO3)2  is mixed with a solution containing 2.25 grams of NaCl. What is the maximum amount of PbCl2 (s) that will form from this reaction?

	19.	At STP, a sample of 35 mL of oxygen is mixed with 20 mL of hydrogen and the mixture ignited. What is the volume, at STP, of any unreacted gas that remains?

	20.	Methane burns in oxygen according to the following equation.
CH4 (g)     +     2 O2 (g)     ®     CO2  (g)      +     2 H2O (g)
A mixture containing 24.0 g CH4 and 128 g O2 is ignited.  How much CO2 is produced?

	21.	What is the limiting reagent when 150.0 g of nitrogen react with 32.1 g of hydrogen?
N(g) + 3H(g)  2NH(g)

	22.	A 500-g sample of Al(SO) is reacted with 450 g of Ca(OH). A total of 596 g of CaSO is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted? 
Al(SO)(aq) + 3Ca(OH)(aq)  2Al(OH)(s) + 3CaSO(s)

Unit 5b Test  Review Sheet
Answer Section

SHORT ANSWER

	1.	ANS:	
13.5 mol of CO; 4.5 mol of NaCHO

PTS:	1

	2.	ANS:	
12.0 mol of HO

PTS:	1

	3.	ANS:	
8.6 L H/ (22.4 L/1 mol)  2 mol HO/2 mol H 22.4 L/ 1 mol = 8.6 L HO

PTS:	1

	4.	ANS:	
3/2

PTS:	1

	5.	ANS:	
15 L of given reactants ´ 11 L of products/10 L of reactants = 16.5 L

PTS:	1

	6.	ANS:	
209.7 g Fe  1 mol Fe/55.85 g Fe  3 mol CO/2 mol Fe  28.01 g CO/1 mol CO = 157.8 g CO

PTS:	1

	7.	ANS:	
45.0 L HS  1 mol HS/22.4 L HS  3 mol O/2 mol HS  22.4 L O/1 mol O = 67.5 L O

PTS:	1

	8.	ANS:	
Theoretical yield:
5.0 g H  1 mol H/2.0 g H  1 mol CHOH/2 mol H  32 g CHOH/1 mol CHOH
= 40 g CHOH
40 g CHOH  86% = 34 g CHOH

PTS:	1

	9.	ANS:	
13.0 L Cl  1 mol Cl/22.4 L Cl = 0.580 mol Cl
103.0 g Na  1 mol Na/23 g Na = 4.48 mol Na 
Cl is limiting reagent: 0.580 mol Cl  2 mol NaCl/1 mol Cl = 1.16 mol NaCl
1.16 mol NaCl  58 g NaCl/1 mol NaCl = 67.3 g NaCl

PTS:	1

	10.	ANS:	
Assume the sodium is limiting: 
48.7 g Na  1 mol Na/23.0 g Na  1 mol H/2 mol Na  (6.02  10 molecules H)/1 mol H 
= 6.37  10 molecules H

PTS:	1

	11.	ANS:	
10.0 L  100%/95% = 10.5 L theoretical yield
10.5 L O  1 mol O/22.4 L O  2 mol KClO/3 mol O  122.6 g KClO/1mol KClO
= 38.4 g KClO

PTS:	1

	12.	ANS:	
8.3 mol HS  2 mol HO/2 mol HS  18 g HO/1 mol = 149.4 g HO theoretical yield
percent yield = 137.1 g/149.4 g  100% = 92%

PTS:	1

PROBLEM

	1.	ANS:	
88.61%

PTS:	1

	2.	ANS:	
Phosphorus is the reactant that is in excess.

PTS:	1

	3.	ANS:	
51.36 g

PTS:	1

	4.	ANS:	
23.56 g

PTS:	1

	5.	ANS:	
18%

PTS:	1

ESSAY

	1.	ANS:	
a) 4Al(s) + 3O2(g) ® 2Al2O3(s)
b) The limiting reagent is the aluminum.
c) 10.3 grams of aluminum oxide can be produced.

PTS:	1

	2.	ANS:	

The molar amounts in this reaction are 9.17 moles nitrogen and 41.5 moles hydrogen, based on the molar masses of each compound. The mole ratio for the entire reaction is 1:3:2. Since 41.5 moles is more than 3 times larger than 9.17 moles, the nitrogen is the limiting reagent. It will only react with 27.5 moles of hydrogen, leaving 14 moles of hydrogen in excess.

PTS:	1

	3.	ANS:	

a) 2 NaOH + Mg(NO3)2 ® 2 NaNO3 + Mg(OH)2.
b) The limiting reagent is NaOH
c) There are 0.1043 moles, or 6.083 grams, of magnesium hydroxide.

PTS:	1

	4.	ANS:	
The equation for the combustion of butane is .
a.	2 molecules C4H10 + 13 molecules O2 ® 8 molecules CO2 + 10 molecules H2O.
b.	2 moles C4H10 + 13 moles O2 ® 8 moles CO2 + 10 moles H2O.
c.	116 g C4H10 + 416 g O2 ® 352 g CO2 + 180 g H2O.
Mass of reactants  Mass of products  532 g.

PTS:	1

	5.	ANS:	
a.	Silver nitrate is the limiting reactant.
b.	The theoretical yield of silver chloride is 38.02 g.
c.	The percent yield of silver chloride is 88.09%.

PTS:	1

	6.	ANS:	
2 KClO3 ®2 KCl + 3 O2
2.50 mol KClO3               =    3.75 mol O2
                                   
 3.75 mol O2    ´            =     120 g O2

PTS:	1

	7.	ANS:	
2 Al + 3 CuCl2 ® 3 Cu + 2 AlCl3

4.30 g Al      0.239 mol Cu

PTS:	1

	8.	ANS:	
CH4 + 2 O2  ® CO2  + 2 H2 O

 4.25 g CO2        6.18 g O2

PTS:	1

	9.	ANS:	
N2 + 3 H2  ®  2 NH3

 6.50 g N2          7.89 g NH3

PTS:	1

	10.	ANS:	
15 g I2  

PTS:	1

	11.	ANS:	
3.00 g Mg3N2  

PTS:	1

	12.	ANS:	
Cu + 2 AgNO3 ® Cu(NO3)2 + 2 Ag

 1.13 g Cu     

PTS:	1

	13.	ANS:	

PTS:	1

	14.	ANS:	
H2 + Cl2  ® 2 HCl

 0.300 mol Cl2    

PTS:	1

	15.	ANS:	
2.50 L H2S   

PTS:	1

	16.	ANS:	
C3H8 + 5 O2 ® 3 CO2 + 4 H2O

200 mL O2   =    0.2 L O2

0.2 L O2

0.12 L CO2   =    120 mL CO2

PTS:	1

	17.	ANS:	
2 Al + 6 HCl ® 3 H2 + 2 AlCl3

35.0 g Al   

PTS:	1

	18.	ANS:	
Pb(NO3)2 (aq) + 2 NaCl (aq) ® PbCl2  (s) + 2 NaNO3

8.00 g Pb(NO3)2      
 
= 6.72 g PbCl2 

  2.25 g NaCl    

 = 5.35 g PbCl2
 
 The maximum amount of PbCl2  that can be formed is 5.35 grams.

PTS:	1

	19.	ANS:	
O2 + 2 H2 ® 2 H2O
35 mL O2        
   
20 mL H2       0.00089 mol H2

Oxygen is in excess.

 0.00089 mol H2   0.00045 mol O2

 0.0016 mol   –   0.00045 mol   =   0.0012 mol O2 in excess

 0.0012 mol      22.4 L   =   0.027 L O2  in excess

 0.027 L        27 mL of oxygen in excess

PTS:	1

	20.	ANS:	
24 g CH4        

128 g O2      

 CH4 is the limiting reactant; 66 grams of CO2 will be produced.

PTS:	1

	21.	ANS:	
150.0 g N  1 mol N/28 g N = 5.36 mol N
32.1 g H  1 mol H/2 g H = 16.1 mol H
5.36 mol N/16.1 mol H = 1 mol N/3 mol H
There is no limiting reagent. The mole ratio of the reactants is exactly 1 mol N to 3 mol H.

PTS:	1

	22.	ANS:	
500 g Al(SO)  1 mol Al(SO)/342 g Al(SO)  3 mol Ca(OH)/1 mol Al(SO)  74 g Ca(OH)/1 mol Ca(OH) = 325 g Ca(OH)
450 g – 325 g = 125 g
There are 125 g excess Ca(OH). Al(SO) is the limiting reagent.
125 g Ca(OH)  1 mol Ca(OH)/74 g Ca(OH) = 1.69 mol Ca(OH)
There are 1.69 mol Ca(OH)2 remaining of unreacted excess reagent.

PTS:	1