Module 2 Problem Set Quiz

Module 2 Problem Set

• Question 1

  Number your answers, and show your work. Be sure to give answers with the correct units.

  Calculate the molecular weight of the following compounds:

  1.  Ca₃N₂             ____________________   

  2.  Li₃PO₄            ___________________

  3.  NH₄NO₃         ____________________

  Correct Answer

  1.  Ca₃N₂ = 3 (40.08) + 2 (14.01) = 148.26

  2.  Li₃PO₄ = 3 (6.941) + (30.97) + 4 (16.00) = 115.793

  3.  NH₄NO₃ = 2 (14.01) + 4 (1.008) + 3 (16.00) = 80.052

• Question 2

  Number your answers, and show your work. Be sure to give answers with the correct units.

  Calculate the number of moles in 10.0 grams in the following compound:

  1.  NH₄NO₃         ____________________

  Calculate the number of grams of each in 0.0500 mol in the following compound:

  2.  Li₃PO₄            ____________________

  Correct Answer

  1. 

  MW of NH₄NO₃ = 80.052

  moles = 10.0 / 80.052 = 0.125 mole 

  2. 

  MW of Li₃PO₄ = 115.793

  grams = 0.0500 x 115.793 = 5.79 grams

• Question 3

  Calculate the percent of each element present in the following compounds:

  1.  Na₂SO₄          Na = __________                S = __________                  O = __________

   2.  Ca(NO₃)₂       Ca = __________                N = __________                  O = __________

  Correct Answer

  PERCENT COMPOSITION

   1.  MW of Na₂SO₄ = 2 (22.99) + (32.07) + 4(16.00) = 142.05

  %Na = 45.98 / 142.05 x 100 = 32.37%

  %S = 32.07 / 142.05 x 100 = 22.58%

  %O = 64.00 / 142.05 x 100 = 45.05%

   2.  MW of Ca(NO₃)₂ = (40.08) + 2 (14.01) + 6 (16.00) = 164.1

   %Ca = 40.08 / 164.1 x 100 = 24.42%

  %N = 28.02 / 164.1 x 100 = 17.07%

  %O = 96.00 / 164.1 x 100 = 58.50%

• Question 4

  Determine the empirical formula for the following compound % compositions:

   1.)  52.56% Fe,    2.83% H,        44.91% O

   2.)  71.98% C,     6.71% H,        21.31% O

  Correct Answer

  1.)  % Composition of a compound is:

  52.56% Fe        52.56% Fe  ÷ 55.85  =  0.941 (smallest number of the set)

  2.83% H            2.83% H  ÷ 1.008    =  2.808

  44.91% O          44.91% O  ÷ 16.00   =  2.807

  0.941 is the smallest of this set of numbers, so it is divided into each of the set of numbers

  Fe  =  0.941  ÷ 0.941  =  1 Fe

  H    =  2.808  ÷ 0.941 =  3 H                       FeH₃O₃

  O   =  2.816  ÷  0.941  =  3 O

  2.)  % Composition of a compound is:

  71.98% C         71.98% C  ÷ 12.01  = 5.993

  6.71% H            6.71% H  ÷ 1.008    = 6.657

  21.31% O          21.31% O  ÷ 16.00   =  1.332 (smallest of the set)

  1.332 is the smallest of this set of numbers, so it is divided into each of the set of numbers

  C = 5.993  ÷ 1.332  = 4.5

  H  = 6.657  ÷ 1.332  = 5

  O  =  1.332  ÷ 1.332  = 1

  Since C is 4.5, multiply all numbers by 2 to get              C₉H₁₀O₂

• Question 5

  BALANCE THE FOLLOWING EQUATIONS:

   1.  SrBr₂     +     (NH₄)₂CO₃     âŸ¶     SrCO₃     +     NH₄Br

   2.  FeS₂     +     O₂     âŸ¶      Fe₂O₃     +     SO₂

  Correct Answer

  1.  SrBr₂     +     (NH₄)₂CO₃     →      SrCO₃     +     2 NH₄Br

   2.  4 FeS₂     +     11 O₂     →      2 Fe₂O₃     +     8 SO₂

• Question 6

  Determine which of the following double replacement reactions will occur using the solubility rules and complete the equations that do:

  1.) Ca⁺², 2 Cl^-1   +   Al⁺³, 3 NO₃^-1   →

   2.) Pb⁺², 2 NO₃^-1   +   2 Na⁺¹, 2 Br^-1   →

  Correct Answer

  1.)  Ca⁺²,  2 Cl^-1   +   Al⁺³, 3 NO₃^-1   →   Will not occur since neither set of ions forms a precipitate

  2.)  Pb⁺², 2 NO₃^-1   +   2 Na⁺¹, 2 Br^-1   →   PbBr₂ ↓   +  2 Na⁺¹, 2 NO₃^-1

• Question 7

  Determine which of the following single replacement reactions will occur using the activity series and complete the equations that do:

   1.)  Cl₂   +   2 F^-1   →

   2.)  Ba   +   Fe⁺²   →

  Balance the following equations and label each reactions as either (1) Combination, (2) Decomposition, (3) Combustion, (4) Double Replacement or (5) Single Replacement.

   3.  C₆H₁₂O₆   +   O₂   →   CO₂   +   H₂0

  4.  HCl   +   NaOH   →   NaCl   +   H₂O

  5.  Al   +   Fe(NO₃)₃   →   Fe   +   Al(NO₃)₃

  6.  S   +   O₂   →   SO₃

  7.  H₂SO₄   →   SO₃   +   H₂O

  Correct Answer

  1.  Cl₂   +   2 F^-1   →   Will not occur since F₂ is more active than Cl₂

  2.  Ba   +   Fe⁺²   →   Fe   +   Ba+2   Will occur since Ba is more active than Fe

  3.  C₆H₁₂O₆   +   6 O₂   →   6 CO₂   +   6 H₂O         Combustion

  4.  HCl   +   NaOH   →   NaCl   +   H₂O                  Double Replacement

  5.  Al   +   Fe(NO₃)₃   →   Fe   +   Al(NO₃)₃             Single Replacement

  6.  2 S   +   3 O₂   →   2 SO₃                                   Combination

  7.  H₂SO₄   →   SO₃   +   H₂O                                  Decomposition

• Question 8

  Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations and then show how those charges are used to balance the following redox equations:

  1.  Mn(NO₃)₂   +   NaBiO₃   +   HNO₃   →   HMnO₄   +   Bi(NO₃)₃   +   NaNO₃   +   H₂O

  2.  NaCrO₂   +   NaClO   +   NaOH    →   Na₂CrO₄   +   NaCl   +   H₂O

  Correct Answer

  1.  Mn(NO₃)₂   +   NaBiO₃   +   HNO₃   →   HMnO₄   +   Bi(NO₃)₃   +   NaNO₃   +   H₂O

  Mn(NO₃)₂: each NO₃ is -1 (total is -2), so Mn is +2

  HMnO₄: H = +1, each O is -2 (total is -8), so Mn is +7

  NaBiO₃: Na is metal in group I = +1, each O is -2 (total is -6), so Bi is +5

  Bi(NO₃)₃: each NO₃ is -1 (total is -3), so Bi is +3

   Since Mn (on left side) is +2 and Mn (on right side) is +7: Mn changes by 5

  Since Bi (on left side) is +5 and Bi (on right side) is +3: N changes by 2

  Multiply Mn compounds by 2 and Bi compounds by 5 and after balancing other atoms =

  2 Mn(NO₃)₂  +  5 NaBiO₃  +  16 HNO₃  →  2 HMnO₄  +  5 Bi(NO₃)₃  +  5 NaNO₃  +  7 H₂O

   2. NaCrO₂   +   NaClO   +   NaOH    →   Na₂CrO₄   +   NaCl   +   H₂O

  NaCrO₂: Na is metal in group I = +1, each O is -2 (total is -4), so Cr is +3

  Na₂CrO₄: Na is metal in group I = +1 (total is +2), each O is -2 (total is -8), so Cr is +6

  NaClO: Na is metal in group I = +1, each O is -2, so Cl is +1

  NaCl: Na is metal in group I = +1, so Cl is -1

  Since Cr (on left side) is +3 and Cr (on right side) is +6: Cr changes by 3

  Since Cl (on left side) is +1 and Cl (on right side) is -1: Cl changes by 2

  Multiply Cr compounds by 2 and Cl compounds by 3 and after balancing other atoms =

  2 NaCrO₂   +   3 NaClO   +   2 NaOH    →   2 Na₂CrO₄   +   3 NaCl   +   H₂O

• Question 9

  Complete the following calculations and show your work. Be sure to give answers with the correct units.

  1.  How many grams of SO₂ would be formed from 60 grams of FeS₂ in the following reaction?

  4 FeS₂     +     11 O₂        →             2 Fe₂O₃     +     8 SO₂

   2.  How many moles of NH₄NO₂ would be required to form 25 grams of N₂ in the following reaction?

  NH₄NO₂           →          N₂     +     2 H₂O

  Correct Answer

  #1

  4 FeS₂  + 11 O₂   →  2 Fe₂O₃  +  8 SO₂

  (60 grams FeS₂ / 119.99) = 0.5000 moles FeS₂

  0.5000 moles FeS₂ x (8 / 4) = 1.00 mole SO₂

  1.00 mole SO₂ x (64.07) = 64.1 grams of SO₂

   #2

  NH₄NO₂   →  N₂  +  2 H₂O

  (25 grams N₂ / 28.02) = 0.8922 moles N₂

  0.8922 moles N₂ x (1 / 1) = 0.892 mol of NH₄NO₂

• Question 10

  Complete the following calculations and show your work. Be sure to give answers with the correct units.

   1. Show the calculation of the mass percent of solute in a solution made by dissolving 15.9 grams of Ca₃(PO₄)₂ in 350 grams of water.

  2. Show the calculation of the molality of a solution made by dissolving 15.9 grams of Ca₃(PO₄)₂ in 400 grams of water.

  3. Show the calculation of the molarity of a solution made by dissolving 15.9 grams of Ca₃(PO₄)₂ to make 350 ml of solution.

  4. Show the calculation of the mass of Ca₃(PO₄)₂ needed to make 200 ml of a 0.128 M solution.

  5. Show the calculation of the volume of 0.238 M solution which can be prepared using 13.4 grams of Ca₃(PO₄)₂.

  Correct Answer

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