Midterm Examination Form IB

James Richard Fromm


Student Name (PLEASE PRINT, FAMILY NAME FIRST)...............................

Student Name (SIGNATURE). . . . . . . . . . . . . . . . . . . . . . .

INSTRUCTIONS

Please enter your name, last or family name first, in the upper right-hand corner of this page, using block printing. Then sign the examination. Time allowed for this examination is 90 minutes. You may use any silent calculator and your own annotated data sheet if you wish.

Answer all questions, placing your answers on the FRONT SIDE of each page, and SHOW YOUR WORK. Use the back sides of pages for scratch paper. Nothing on the back side of a page will be graded.


1. (10 pts) Name, using correct modern chemical nomenclature, each of these compounds: a. MgO; b. HF; c. KHCO3; d. Ca3(PO4)2; e. H2SO3; f. HClO4; g. KBr; h. NaBrO; i Fe2O3; j. NiCl2

2. (10 pts) Give the correct molecular formula for each of the following compounds. If no molecular formula exists, give the correct empirical formula: a. hydrochloric acid; b. nickel(II) carbonate; c. potassium nitrate; d. iron(III) sulfate; e. sodium iodide dihydrate; f. calcium fluoride; g. lithium oxide; h. sodium hydrogen sulfate; i. potassium iodate; j. hypochlorous acid

3. (15 pts) Calculate the percentage composition by mass of anhydrous sodium dihydrogen phosphate, NaH2PO4.

4. (10 pts) Calculate the percent yield of a process which produces 36 g of metallic nickel from 148 g of nickel(II) sulfide, NiS.

5. (10 pts) A chemist produced the toxic gas hydrogen sulfide, H2S, by adding acid to solid iron(II) sulfide, FeS. If he started with 0.020 moles of FeS and converted all of it to H2S, calculate the partial pressure of the H2S at 25oC in a volume of 50,000 dm3 (the approximate volume of a room).

6. In an industrial plant, carbon is burned to carbon monoxide. The carbon monoxide is used to reduce iron ore (Fe2O3), producing iron metal and carbon dioxide.
a. (2 pts) Write the complete balanced equation for the reduction of Fe2O3 with CO.
b. (6 pts) Calculate the heat absorbed or evolved in this reduction, using the equation you wrote above, the standard enthalpy of formation of Fe2O3 as -824.2 kJ/mol, and appropriate other thermodynamic data.
c. (2 pts) The temperature to which this calculation above applies is __________, and at that temperature the reaction is (endothermic, exothermic).

7. A chemist added 96.42 mL of 0.0463 molar aqueous sodium hydroxide solution to 11.35 mL of 0.0591 molar aqueous hydrochloric acid solution. The reaction OH-(aq) + H+(aq) --> H2O took place immediately and quantitatively. The ions Na+(aq) and Cl-(aq) did not react with anything.
a. (4 pts) Calculate the molar concentrations of Na+(aq) and Cl-(aq) in the final solution.
b. (6 pts) After the reaction takes place, the concentration of one of the two reacting ions will be nearly zero. The ion of these two which will remain in significant concentration is (hydroxide ion, hydrogen ion). The molar concentration of this ion should be ____________________.

8. An industrial reaction of considerable interest is the formation of methane from coal. For the reaction 2C(s) + 2H2O(g) --> CH4(g) + CO2(g), the equilibrium constant at 25oC and one bar has the value of 195 with no nominal units.
a. (5 pts) Write the correct form of the equilibrium constant for this reaction.
b. (10 pts) Compute the partial pressures of methane and of carbon dioxide at equilibrium when the partial pressure of water vapor (steam) is held at 1.00 bar. [Note: Stay with the pressure unit of bar, 100,000 Pa, to do this calculation most readily.]

9. Consider the dimerization of nitrogen dioxide, 2NO2(g) <--> N2O4(g).
a. (2 pts) An increase in total pressure should (increase relative p(N2O4), increase relative p(NO2), have no significant effect).
b. (4 pts) The reaction is (endothermic, exothermic) at 25oC. Therefore, at or near 25oC, an increase in temperature should favor the production of (NO2,N2O4).
c. (4 pts) A supply of additional NO2(g) is injected into this gas mixture at equilibrium. The resulting partial pressure of NO2(g) (is, is not) the same as the original equilibrium partial pressure of NO2(g) plus the partial pressure of the added NO2 because:


Answer Key

1. a, manganese(II) oxide (the II is optional); b, hydrogen fluoride or hydrofluoric acid; c, potassium hydrogen carbonate; d, calcium phosphate; e, hydrogen sulfite or sulfurous acid; f, hydrogen perchlorate or perchloric acid; g, potassium bromide; h, sodium hypobromite or, in I.U.P.A.C. form, sodium monoxochlorate(I); i, iron(III) oxide or, less commonly, diiron triiodide; j, nickel(II) chloride or nickel dichloride.

2. a, HCl; b, NiCO3; c, KNO3; d, Fe2(SO4)3; e, NaI.2H2O; f, CaF2; g, Li2O; h, NaHSO4; i, KIO3; j, HClO.

3. 1 x 23.00 = 23.00 g = mass Na
2 x 1.01 = 2.02 g = mass H
1 x 30.97 = 30.97 g = mass P
4 x 16.00 = 64.00 g = mass O
molar mass = 119.99 g
100 x 23.00/119.99 = 19.2% Na by mass
100 x 2.02/119.99 = 1.7% H by mass
100 x 30.97/119.99 = 25.8% P by mass
100 x 64.00/119.99 = 53.3% O by mass

4. molar mass of NiS = 58.69 + 32.06 = 90.75 g/mol
148 g NiS/90.75 g/mol NiS = 1.631 mol NiS
NiS forms Ni, 1 mol NiS produces 1 mol Ni
mass Ni expected = 1.631 mol Ni x 58.69 g/mol Ni= 95.715 g Ni
100 x 36 g/95.715 g = 37.6% yield (as Ni)

5. FeS(s) + 2H+(aq) --> Fe2+(aq) + H2S(g)
0.020 mol FeS --> 0.020 mol H2S
pV = nRT, p = nRT/V
p = (0.020 mol)(8.314 kPa dm3 mol-1 K-1)(298.15 K)/50,000 dm3
p = 9.92 x 10-4 kPa = 9.92 x 10-1 Pa = 0.992 Pa

6. a. 2Fe2O3(s) + 6CO(g) --> 4Fe + 6CO2(g)
b. DHoR = S DHof (products) - S DHof (reactants)
DHoR = 4(0) + 6(-394) - [2(-824.2) + 6(-111)] kJ
DHoR = -2364 - [-1648.4 + (-666)] kJ
DHoR = -2364 - [-2314.4] = -49.6 kJ/mole reaction as written above.
If you had chosen to write Fe2O3 + 3CO --> 2Fe + 3CO2, you would have obtained half the above answer for DHoR.
c. 298.15 K (25oC); exothermic
7. a. final volume is 96.42 + 11.35 = 107.77 mL
initial mol Na+ = 96.42 mL x 0.0463 mmol/mL = 4.46 mmol = final mol Na+, since Na+ does not react.
concentration Na+ = 4.46 mmol/107.77 mL = 0.0414 M
initial mol Cl- = 11.35 mL x 0.0591 mmol/mL = 0.67 mmol = final mol Cl-
concentration Cl- = 0.67 mmol/107.77mL = 0.00622 M
b. hydroxide ion
original mol OH- = mol Na+ = 4.46 mmol
original mol H+ = mol Cl- = 0.67 mmol
Reacting in a 1:1 ratio, the limiting reagent is the hydrogen ion, so hydroxide will remain in significant (left-over) concentration while hydrogen ion will not.
4.46 - 0.67 = 3.79 mmol hydroxide ion left over.
Its molar concentration is then 3.79 mmol/107.77 mL = 0.0352 mol/L.

8. a. Kp = p(CH4) p(CO2/p2(H2O)
The activity of carbon is taken as one because it is a pure solid substance.
b. 195 = p(CH4) p(CO2)/p2(H2O)
p(H2O) = 1 bar
p(CH4) = p(CO2) from the reaction stoichiometry
195 = p2(CH4)/1.0 bar2
195 bar2 = p2(CH4), p(CH4) = p(CO2) = 13.96 bar
The equilibrium will favor CH4(g) and CO2(g) at 25oC.


Copyright 1997 James R. Fromm