1) 27.6 g K₂CO₃ = 0.200 mol K₂CO₃

0.200 mole K₂CO₃ = 0.0167 mol K₂Zn₃[Fe(CN)₆]

0.0167 mol K₂Zn₃[Fe(CN)₆] = 11.6 g K₂Zn₃[Fe(CN)₆]

2) C_xH_y

3.30 g CO₂ = 0.075 mol C

0.899 g H₂O = 0.050 mol H₂O = 0.100 mol H

C₁H_1.33.... or C₃H₄

Addendum 9/26/99: The empirical and molecular formula are the same.

3) C₂H₇N
Addendum 9/26/99: According to Avogadro's Hypothesis, equal volumes of gases at the same temperature and pressure contain an equal number of molecules of gas. 1 volume = 1 mole in the equation that contains all gaseous substances. Empirical formula and molecular formula are the same.

4) Let x = g of Cu₂O

And y = g of CuO

Then x + y = 1.000

.888(1 - y) + y(0.799) = 0.839

y = 0.55.. or 55% CuO... or 0.55 g in a 1.000 gram sample.

5) Since 1 BaBr₂ = 1 BaCl₂

Solving for at. wt. of Ba gives 137

6) Let x = weight of KBr and let y = weight of NaBr

x + y = 0.560 gram

Weight of AgBr from KBr = 1.57x since

Weight of AgBr from NaBr = 1.82y since

1.57x + 1.82y = 0.970

Solve two equations in x and y simultaneously

Gives KBr = x = 0.197

 Lightbulb/Tungsten Filament Problems

7) The commercial isolation of tungsten begins with the mineral Scheelite, the principal component of which is calcium tungstate (CaWO₄), which is in turn converted to tungstic acid, the hydrous (hydrated) oxide (H₂WO₄), by treatment with concentrated hydrochloric acid (HCl). At elevated temperatures, tungstic acid can be reduced in a charcoal lined furnace directly to the pure metal. (For helpful information, see the discussion of the incandescent light bulb and the tungsten filament on the handout.)

(a) Write a balanced chemical equation for the reduction step in which tungstic acid is reduced to the metal.

H₂WO₄ + heat = H₂0 + WO₃

WO₃ + 3H₂ = W + 3H₂O

(b) In the key reduction step, a 100. gram sample of pure tungstic oxide is reduced (as described above) with 10.0 grams of charcoal. What is the limiting reagent and the maximum yield (in grams) of tungsten that can be realized by this process? Assume charcoal to be pure carbon.

2WO₃ + 3C = 2W + 3CO₂

Limiting reagent is tungstic oxide.

Maximum yield of W is 79g.
Addendum 9/26/99: 7.76 g carbon is needed for 100 g of WO3. You have 10 g so carbon is in excess and WO3 is limiting.

8) A mixture of calcium tungstate (CaWO₄) and iron tungstate (FeWO₄) weighed 1.0 grams. Both of these insoluble tungstates were converted to the soluble sodium tungstate (Na2WO4) which also weighed 1.0 grams. Determine the percent calcium tungstate in the mixture, assuming only those two substances are present.

Addendum 9/26/99: The answer is 38.5% FeWO₄, not 33.3%.

9) To estimate the number of tungsten atoms in a 15-watt light bulb filament, see the incandescent light bulb discussion in the handout. Here are two ways to approach this problem, both of which should give you an order-of-magnitude result:

First: Using the statement that 1 lb of tungsten light bulb filament produces 200 miles of filament and that 30 inches of this filament is needed for a 15 watt bulb gives a result of about 4x10 to the 18th atoms.

Second: Using the statement that the tungsten filament used in the bulb is 1/100mm or about 0.0004 inches in diameter, assuming the filament is a solid rod of uniform diameter and known length, and estimating the density of tungsten to be about 18 or 19 g per cm³ gives and order-of-magnitude calculation of the number of atoms of about 10 to the 19th.