Introductory Chemistry Midterm Review

Sorry about the poor formatting. If you don’t understand something, let us know!

Question 10: ENTHALPY OF FORMATION

In a thermite reaction, iron oxide reacts with aluminum metal:

3Fe3O4(s) + 8Al(s) → 9Fe(s) + 4Al2O3(s)

Using the provided data, calculate the change enthalpy of formation of the reaction.

The change in enthalpy formation(kJ/mol) of selected substances:

Al(g): 330
Al2O3(s): -1676
Fe(g): 416
Fe3O4(s): -1117

3(-1117 kJ) + 8(0) + 4(-1676 kJ)
-3351 kJ -6704 kJ

Change in enthalpy formation = -6704 kJ – (-3351 kJ)
= -3353 kJ (exothermic)

Question 9: LIMITING REAGENTS

N2(g) + H2(g) → NH3(g)

In a stainless steel container, 84.0 g of N2(g) reacts with 36.0 g of H2(g)

(a) Balanced chemical equation: N2(g) + 3H2(g) → 2NH3(g)

Note: Order of Balancing: CHO

(b) Calculate the number of moles of N2(g) and the number of moles of H2(g).

36.0 g H2 (1 mol H2/2.02g H2) = 17.8 mol H2

84.0 g N2 (1 mol N2/28.01 g N2) = 3.00 mol N2

(c) Limiting Reagent: 17.8 mol H2 (2 mol NH3/3 mol H2) = 11.9 mol NH3

3.00 mol N2 (2 mol NH3/1 mol N2) = 6.00 mol NH3

THEREFORE, the limiting reagent is NITROGEN

(d) Theoretical Yield: 3.00 mol N2 (2 mol NH3/1 mol N2)(17.04 g NH3/1 mol NH3) = 102 g NH3

(e) Percent yield, given an actual yield of 75.6 g:

REMEMBER, formula for percent yield = (actual/theoretical) x 100 = (75.6g/102g) x 100 = 74.1%

NOTE: EXCESS REAGENT:

17.8 mol H2 and 3.00 mol N2

Find the amount that is used up → 9.00 mol H2 have been used = 8.8 mol of H2 remain

8.8 mol H2 (2.02 g H2/1 mol H2) = 18 g H2 of excess reagent

Question 8: Oxidation Numbers

Note: If H+ is with a non-metal it is positive, and if it is with a metal H+ will be negative.

(a.) Mn in MnO2- → x + 2(-2) = -1 → x = +3

(b.) B in BH3 → x + 3(-1) = 0 → x = +3

(c.) Cl in ClO4- → x + (-2)(4) = -1 → x = +7

(d.) Ca in CaF2 → x +2(-1) = 0 → x = 2

(e) HBrO3 → x + 3(-2) = 0 → x = 5

7. Oxidation-Reduction Reactions

a. Ca(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2(g) Calcium is oxidized, Hydrogen is reduced

0 +1 -1 +2 0

b. 2Ag+(aq) + Mn(s) → Mn 2+(aq) + 2Ag(s) Mangenese is oxidized, and silver is reduced

+1 0 +2 0

6. Molarity

REMEMBER → M1V1 = M2V2

REMEMBER → Strong Acids and Strong Bases

a) (10.0 mol/L)(x) = (0.500 mol/L)(0.450 L)

x = [(0.450 L)(0.500 mol)]/(10.0 mol/L)

x = 0.0225 L (1000 mL/1 L) = 22.5 mL HNO3

b) (0.500 mol/L HNO3)(0.400 L HNO3)(1 mol KOH/1 mol HNO3)(1 L/0.200 mol KOH) = 1.00 L(1000 mL/1 L) = 1.00 x 10^ 3 mL

5. Bond Enthalpies

H H H

| | |

H-C-C-C-H + 5 O=O → 3 O=C=O + 4 H-O-H

| | |

H H H

8 CH 2 CC 5(O=O) 3(2(O=C)) 4(2(OH))

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

C-H:413 kJ/mol C=O: 799 kJ/mol

O=O: 495 kJ/mol H-O: 463 kJ/mol

C-C: 348 kJ/mol

8(413 kJ) + 2(348 kJ) + 5(495 kJ) – ((6(799 kJ)) + 8(463 kJ))

Bond Enthalpy = -2023 kJ (exothermic reaction)

4. Writing Chemical Equations:

Solid magnesium oxide reacts with aqueous hydrochloric acid to form aqueous magnesium chloride and liquid water.

a. Molecular Equation: MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)

b. Complete Ionic Equation: MgO(s) + 2H+(aq) + 2Cl-(aq) → Mg 2+(aq) + 2Cl-(aq) + H2O(l)

c. Spectator Ions: 2Cl-(aq)

d. Net Ionic Equation: MgO(s) + 2H+(aq) → Mg 2+ (aq) + H2O(l)

3. Calorimetry

When a 3.53 g sample of KOH(s) dissolves in 0.100 kg of H2O in a coffee-cup calorimeter, the temperature rises from 21.1 C to 27.4 C. The heat capacity of the calorimeter apparatus is 145 J/K. Calculate the enthalpy of the reaction for the dissolution of KOH:

KOH(s) → K+(aq) + OH-(aq)

Assume the specific heat of the solution is 4.18 J/K x g

In this case, add the sample plus water for mass

qrxn = -[(smT) + (CcalT)]

= [(4.18 J/K x g)(103.53g)(6.3 K)) + (145 J/K)(6.3 K)

= -3600 J (1kJ/1000 J)

= -3.6 kJ

2. Bomb Calorimetry

Acetylene is a gas used in welding. A 1.00-g sample of C2H2 was burned in a bomb calorimeter who total heat capacity is 24.2 kJ/K. The temperature of the calorimeter apparatus increased by 4.12 K. Calculate the heat of combustion per mole of acetylene.

Qrxn = -[CcalT]

= -[(24.2 kJ/K)(412 K)]

= -99.7 kJ

BUT WE ARE NOT DONE! …

-99.7 kJ/1.00g (26.04 g C2H2/1 mol C2H2)

= -2.60 x 10^3 kJ/mol

1. Moles

Suppose you have a 1.00 M MgCl2 solution.

a. Calculate the moles of MgCl2 in 200 mL of the solution.

1.00 mol/L MgCl2 (200 mL MgCl2)(1L/1000 mL) = 0.2 mol MgCl2

b. How many moles of Cl- are present in 1.00 L of the MgCl2 solution?

MgCl2 (aq) → Mg2+ (aq) + 2Cl- (aq)

1.00 mol/L MgCl2 (1.00 L)(2 Cl-/1 MgCl2) = 2.00 mol Cl-