# Chapter 4: Chemical Quantities and Aqueous Reactions

## 4.2 Reaction Stoichiometry: How Much Carbon Dioxide

1. Calculate the mass of phosphorous trichlorided produced when 0.01 moles of P4 reacts with 0.06 moles of chlorine gas according to the balanced equation.

P4 (s) + 6 Cl2 (g) -> 4 PCl3

 P4 (s) 6 Cl2 (g) -> 4 PCl3 (l) Given (moles) 0.01 mole 0.06 mole Use 0.01 mole 0.06 mole Produce 0.04 mole Final 0 mole 0 mole 0.04 mole MW 123.895 g mole-1 70.905 g mole-1 137.332 g mole-1 Given (mass) 1.24 g 4.25 g Produce (mass) 5.49 g

## 4.3 Limiting Reactant, Theoretical Yield, and Percent Yield

1. Calculate the amount of product and reactant remaining in the video clip with the reaction of zinc metal and hydrochloric acid. ( Internet© Saunders, 1997). Assume all three use 1.0 moles of HCl, and that the flasks contain 1.0, 0.5 and 0.25 moles of zinc metal.
2.  Zn (s) 2 HCl (aq) --> Zn2+ H2 (g) Initial 1.0 1.0 Change 0.5 1.0 0.5 0.5 Final 0.5 0 0.5 0.5

 Zn (s) 2 HCl (aq) --> Zn2+ H2 (g) Initial 0.5 1.0 Change 0.5 1.0 0.5 0.5 Final 0 0 0.5 0.5

 Zn (s) 2 HCl (aq) --> Zn2+ H2 (g) Initial 0.25 1.0 Change 0.25 0.5 0.5 0.5 Final 0 0.25 0.25 0.25

3. Calculate the amount of product and reactant remaining when the following react according to the balanced chemical equation:
2 H2 + O2 -> 2 H2O
2. Calculate limiting reagent
3. Calculate final values
4. If only 10.58 g H2O what is the % yield
5.  2 H2 O2 -> 2 H2O given (mass) 10.0 g 10.0 g MW 2.016 g mole-1 31.999 g mole-1 18.015 g mole-1 given (mole) 4.96 mole 0.312 mole use 0.625 mole 0.312 mole produce 0.625 mole final (mole) 4.335 mole 0 mole 0.625 mole final (mass) 8.74 g 0 g 11.26 g

4. Solid Rocket fuel used for the space shuttle is aluminum metal and ammonium perchlorate. These react to produce aluminum oxide, aluminum chloride, nitrogen monoxide, and water (SRB info from NASA).
1. Write out the chemical equation
2. Balance the chemical equation
3. Given 1.00 kg each reactant, which is limiting, what is the mass of each product.
4. Given 1.00 kg Al, how much NH4ClO4 should be used. What is the mass of products?
5. 3 Al(s) + 3 NH4ClO4(s) -> Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)
6.  3 Al (s) 3 NH4ClO4 (s) -> Al2O3 (s) AlCl3 (s) 3 NO (g) 6 H2O (g) given (gram) 1.00x103 1.00 x103 MW (g mole-1) 26.9815 117.4886 101.961 133.3396 30.0061 18.0152 given (mole) 37.06 8.51 use (mole) 8.51 8.51 produce (mole) 2.84 2.84 8.51 17.02 final (mole) 28.55 0 2.84 2.84 8.51 17.02 final (gram) 770. 0 290. 379 255 307

5. Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors (video clip from "The Chemistry Set") What masses of iron(III) oxide and aluminum are required for a theoretical yield of 15.0 g iron? What is the maximum mass of aluminum oxide that could be produced? How much aluminum oxide is actually produced from these starting materials if the yield is 93%? The thermite reaction is:
6. Fe2O3(s) + 2Al(s) -> 2Fe(l) + Al2O3(s)

 Fe2O3(s) 2Al(s) -> 2Fe(l) Al2O3(s) produce 15 g MW 159.7 g/mole 26.98 g/mole 55.85 g/mole 102.0 g/mole Produce 0.269 mole 0.134 mole Need 0.134 mole 0.269 mol Need 21.4 g 7.26 g Yield (theoretical) 15 g 13.7 g 93% Yield 14.0 g 12.7 g

7. Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz. Balance the following reaction.
1. What mass of C6H8O7 should be used for every 1.0*102 mg of NaHCO3?
2. If 1.0*102 mg of C6H8O7 is used, which is the limiting reagent?
3. What mass of CO2(g) could be produced by this mixture?
4. If only 45 mg of CO2 is produced, what is the percent yield?
5. How many grams of Na3C6H5O7 is actually produced?
6.  3 NaHCO3(aq) 1 C6H8O7 -> 3 CO2 3 H2O 1 Na3C6H5O7(aq) given 1.0*102 mg given 0.10 g MW 84.01 g/mole 192.1 g/mole 44.01 g/mole 18.02 g/mole 258.07 g/mole given 1.19x10-3 mole need 3.97x10-4 mole need 0.076 g produce 1.19x10-3 mole 1.19x10-3 mole 3.97x10-4 mole Yield (theoretical) 0.052 g 0.021 g 0.102 g Actual Yield 0.045 g % yield 86.5 % Actual Yield 0.018 g 0.088 g

## 4.4 Solution Concentration and Solution Stoichiometry

1. NiCl2*6H2O ( internet)
1. 8.320g NiCl2*6H2O
3.  Mass of Compound (NiCl2*6H2O) 8.320 g MW (NiCl2*6H2O) 237.69 g mole-1 Moles (NiCl2*6H2O) 0.0350 mole Volume (NiCl2*6H2O) 250.0 mL Concentration (NiCl2*6H2O) 0.140 mole liter-1 What happens to ions? internet Concentration Ni2+ 0.140 mole liter-1 Concentration Cl1- 0.280 mole liter-1

2. How to produce 500.0 mL of 0.0100 M K2Cr2O7 (aq)
3.  Concentration (K2Cr2O7 (aq) ) 0.0100 M Volume (K2Cr2O7 (aq) ) 500.0 mL Moles (K2Cr2O7 (aq) ) 0.00500 mole MW (K2Cr2O7 (aq) ) 294.18 g mole-1 mass (K2Cr2O7 (aq) ) 1.471 g Concentration K1+ 0.0200 M Concentration Cr2O72- 0.0100 M

4. How do you prepare 100 mL of 0.150 M caffeine (C8N4O2H10)?
5. How do you prepare 1.50 L of 3 M Mg2+ from Magnesium nitrate?

### Dilution

1. 0.100 M K2Cr2O7 2.00 mL diluted to 500 mL
1. ( internet )
2.  Initial Concentration (K2Cr2O7 (aq) ) 0.100 mole liter -1 Volume (K2Cr2O7 (aq) ) 2.00 mL Moles (K2Cr2O7 (aq) ) 0.000200 mole Dilution Volume 500 mL Final Concentration (K2Cr2O7 (aq) ) 0.000400 mole liter-1

2. 0.140 M NiCl2 (aq) solution
1. Dilute 2.00 mL to 500 mL
2.  Initial Concentration NiCl2 0.140 mole liter-1 Volume NiCl2 2.00 mL Moles NiCl2 2.8x10-4 mole Final Volume NiCl2 500 mL Final Concentration NiCl2 5.6x10-4 mole liter-1

3. Dilute 5.00 mL to 500 mL
4.  Initial Concentration NiCl2 0.140 mole liter-1 Volume NiCl2 5.00 mL Moles NiCl2 7.0x10-4 mole Final Volume NiCl2 500 mL Final Concentration NiCl2 1.4x10-3 mole liter-1

5. Dilute 5.00 mL to 250 mL
6.  Initial Concentration NiCl2 0.140 mole liter-1 Volume NiCl2 5.00 mL Moles NiCl2 7.0x10-4 mole Final Volume NiCl2 250 mL Final Concentration NiCl2 2.8x10-3 mole liter-1

7. Dilute 5.00 mL to 1.00 L
8.  Initial Concentration NiCl2 0.140 mole liter-1 Volume NiCl2 5.00 mL Moles NiCl2 7.0x10-4 mole Final Volume NiCl2 1.00 L Final Concentration NiCl2 7.0x10-4 mole liter-1

## 4.8 Acid-Base and Gas-Evolution Reactions

1. Magnesium Hydroxide + Nitric Acid
1. Balanced total equation: Mg(OH)2 (aq) +2 HNO3(aq)-> 2 H2O +Mg(NO3)2 (aq)
2. Total Ionic Equation: Mg2+ + 2 OH1- + 2 H1+ + 2NO31- - > 2H2O + Mg2+ + 2 NO31-

3. Net Ionic Equation 2 OH1- + 2 H1+ -> 2H2O

2. Sulfuric acid and lithium hydroxide
1. Total equation: H2SO4 (aq) + 2 LiOH (aq) -> 2 H2O + Li2SO4 (aq)

2. Ionic Equation: 2 H1+ (aq) + SO42- (aq) + 2 LI1+ (aq) + 2 OH1- -> 2 H2O + SO42- (aq) + 2 Li1+ (aq)

3. Net ionic equation 2 H1+ (aq) + 2 OH1- -> 2 H2O

3. Acetic acid and iron (III) hydroxide
1. Total equation: 3 CH3COOH (aq) + Fe(OH)3 (s) -> 3 H2O + Fe(CH3COO)3 (aq)

2. Ionic Equation: 3 CH3COO1- + 3H1+ (aq) + Fe3+(aq) + 3 OH1- (aq) ->
3 H2O + Fe3+ (aq) + 3 CH3COO1- (aq)

3. Net ionic equation: 3H1+ (aq) + 3 OH1- (aq) -> 3 H2O

4. Ammonia (NH3) and water( Internet)
NH3 + H2O -> NH41+ (aq)+ OH1-(aq)

5. hydrochloric acid and ammonia produces ammoinum and chloride
HCl (aq) + NH3(aq) - > NH41+ (aq) + Cl1-(aq)

6. hydrochloric acid and sodium hydroxide produces sodium chloride and water; no indicator
HCl(aq) + NaOH(aq)-:gt; H2O + NaCl(aq)