From WolfWikis

Exercises

Exercise 1

Two moles of hydrogen and two moles of oxygen are brought together in a balloon and ignited.

1. How much water is formed

   Need the balanced equation

   2H₂ + O₂ ==> 2H₂O

   I think you can already solve this by having a good look at the equation!

   Instead of the balanced ratio 2:1::2 we have 2:2 available, clearly there is excess of oxygen and hydrogen is limiting!

2. Which reactant is in excess

   oxygen

3. How much of it is left after the reaction has gone to completion

   As only one mole of oxygen can react, one mole will be left watching the dance.

Exercise 2

Consider the reaction Ca + H₂O ==> H₂ + Ca(OH)₂

One gram of calcium is brought into 6 liters of water

1. What is the limiting reactant?

   Hmmm, one lousy gram of Ca and a bucket full of water? I wonder what is limiting...

   But let's go through the motions anyhow, shall we? First balance that equation already.

   Ca + 2H₂O ==> H₂ + Ca(OH)₂

   In the improbable event that water is limiting how much hydrogen could we get. Let's go to moles.

   6 liters equals 6000 ml equals 6000 cm³.

   Using its density of 1 [g/cm³] multiplication gives 6000 g

   Molar mass for water is 16+2=18 [g/mol].

   Division gives 6000/18 [g][mol/g] = 333.33 [mol]

   The equation tells us that for every two water molecules we only get one hydrogen, so we would get 333.33/2 = 166.67 [mol] hydrogen

   Now start from calcium.

   1 [g] Ca, M= 40 [g/mol].

   Division gives 1/40 [g][mol/g] = .025 [mol] of Ca that would produce .025 [mol] of hydrogen

   What would be smaller? I think it is indeed the calcium that is in short supply.

   Unsurprisingly.

2. What volume of hydrogen is formed (at 1 Atm and 25^oC)

   As for each mole of Ca we get one mole of hydrogen, we should be getting 0.025 [mol]

   At these conditions the molar volume is 22.4 [l/mol]

   Multiplication gives 22.4*0.025 [l/mol][mol] = 0.56 [l]

   .

   As you can see that is still a fair bit. Though it can be very useful to have a gut feeling of how a calculation might fall out, because that makes you spot simple computational errors, calculations can sometimes tell you your gut feelings were wrong too...

Exercise 3

Gold can react with chlorine to form AuCl₃

1. How much of the excess reactant is left (by weight) if we start with 1 gram of gold and 2 liters of chlorine (at 1 Atm and 25^oC)?

   1 g of Au, M = 197 [g/mol]. Division gives 1/197 [g][mol/g] = 5.07 [mmol] of Au

   2 l of Cl₂, molar volume 22.4 [l/mol]. Division gives 2/22.4 [l][mol/l] = 89.3 mmol

   Balancing:

   2Au + 3Cl₂ ==> 2AuCl₃

   If I divide the number of available mmoles of Au (5.07) by its coefficient (2) I get 2.535

   Multiplying the whole equation with that number I find

   5.07Au + 7.605 Cl₂ ==> 5.07AuCl₃

   So the 5.07 mmoles of available gold can only react with a little more than 7.6 mmoles of chlorine, not the 89.3 mmoles that are available.

   Clearly Au is limiting

   What's left of the chlorine is 89.3-7.6= 81.7 [mmol].

   The molar mass is 2*35.5=71 [mg/mmol]

   Multiplying we find 71*81.7 [mg/mmol][mmol] = 5800 [mg] = 5.8 [g] of chlorine left.

remark: we heat up the mixture to an appropriate temperature of 200^oC to facilitate the reaction.

1. Nice try, but this remark was pretty irrelevant. Yes the gas will expand (if it is kept at 1 Atm) or get a higher pressure (if the volume is kept the same) but the number of atoms will not be affected, as long as we make sure none of the gas can escape or the gold stolen.

Exercise 4

Combustion reactions are typically carried out in bomb calorimeters. A pellet of the substance under investigation is brought into a sturdy metal container (the bomb) with an half-inch thick wall and a equally sturdy lid that can be screwed on. The bomb is pressurized with 25 Atm of pure oxygen and a current is yanked through a little fuse wire that rests on top of the pellet. The wire glows red hot and ignites the reaction. The idea is that the oxygen is in excess, that the pellet burns up completely without generating soot. The heat evolved can be measured by putting the bomb in a vat with a lot of water and the temperature change of the water is measured with a good thermometer.

Interesting story and all true too, but most of it is pretty irrelevant for the problem

The volume inside the bomb is 500 ml and the oxygen pressure 25 atm and the pellet is 6 grams of solid benzoic acid (C₆H₅COOH)

1. This is much more interesting for the calculation.

2. Determine the limiting reactant.

   When in doubt balance an equation.

   C₆H₅COOH is the same as C₇H₆O₂ and that's really all we need for the equation.

   Besides it is a combustion of an organic so we know the only products are water and carbon dioxide.

   2C₇H₆O₂ + 15O₂ ==> 14CO₂ + 6H₂O

   So the ratios are 2:15::14:6

   At room temperature and 25 Atm the molar volume would be 22.4/25 [l/mol] = .896 [l/mol]

   This is so because PV=1RT for n=1 and RT is constant, so increasing P by a factor of 25 decreases the molar volume by the same factor.

   Thus we have 0.5 [l] divided by 0.896 [l/mol] equals 0.55 [mol] of oxygen

   The molar mass of benzoic acid is 7*12+6*1+2*16 = 122 [g/mol]

   Dividing 6 [g] by 110 [g/mol] we find 0.04912 mol of benzoic acid

   This means there is more than ten times more oxygen than bezoic, but the ratio required is only 15:2 = 7.5.

   Oxygen is in excess although not terribly so.

3. How much water is formed?

   All the benzoic will burn and produce water in a ratio of 2::6 or 1::3. Thus we get 3*.04912 = 0.148 [mol] of water.

4. How much benzoic acid must be used for the stoichiometries of the reactants to match?

   As the balanced ratio calls for 2:15, i.e. a factor of 7.5 we should divide the available 0.55 [mol] by a factor of 7.5.

   This gives .0733 [mol] or 8.946 [g] of benzoic acid.

   Obviously we should not make that pellet too big, otherwise we end up with a sooty mess and a bad measurement.

Exercise 5

CZX1 is a compound^[1] that can be synthesized from zinc chloride ZnCl₂, copper(I)chloride (CuCl) and trimethylammoniumchloride (TMACl) (CH₃)₃NHCl, all anhydrous. Every sixth zinc atom is replaced by the combination of a copper atom plus a TMA moiety, leading to a balanced reaction equation:

5ZnCl₂ + 1CuCl + 1TMACl ==> 1CZX1^[2]

1. Starting with 10g of zinc chloride, 3g of CuCl and 3g of TMACl, what is the limiting reactant

   The ratios are 5:1:1::1

   We need molar masses

   ZnCl₂ = 65.4+2*35.5 = 136.4 [g/mol]

   CuCl =63.5+35.5 = 99 [g/mol]

   (CH₃)₃NHCl = C₃H₉NHCl ==> =3*12+9*1+14+1+35.5= 95.5 [g/mol]

   Dividing by mola masses we get moles

   ZnCl₂ = 10 [g]/ 136.4 [g/mol] = 0.2199 [mol]

   CuCl = 3 [g] /99 [g/mol] = 0.10101 [mol]

   C₃H₉NHCl = 3 [g]/ 95.5 [g/mol] = 0.0314

   To calculate the number of moles of product, all we need to do is divide the zinc chloride number of moles by 5 because the other coefficients are 1

   product based on: ZnCl₂ = 10 [g]/ 136.4 [g/mol] = 0.02199 [mol]/5 = 0.004399 [mol]

   product based on: CuCl = 3 [g] /99 [g/mol] = 0.10101 [mol]

   product based on: C₃H₉NHCl = 3 [g]/ 95.5 [g/mol] = 0.0314 [mol]

   Clearly the zinc chloride is limiting the reaction.

2. How how much CZX1 is formed?

   4.399 [mmol]

Notes

1. ↑ This compound was the topic of Amanda Josey's recent thesis. Dr. Josey worked under Dr. J. Martin and is now working at BASF.
2. ↑ Actually: Zn₅Cu[(CH₃)₃NH]Cl₁₂. It is not unusual if compounds get this complicated to find a suitable abbreviation for them.