o
- Reacting mass calculation Example 6a.5
o The production of copper from a copper ore
o (a) Theoretically how much copper can be obtained from 2000 tonne of pure chalcopyrite ore, formula CuFeS2 ?
o Chalcopyrite is a copper-iron sulphide compound and one of the most important and common ores containing copper.
o Atomic masses: Cu = 64, Fe = 56 and S = 32
o For every one CuFeS2 ==> one Cu can be extracted, formula mass of ore = 64 + 56 + (2x32) = 184 o Therefore the reacting mass ratio is: 184 ==> 64
o so, solving the ratio ...
2000 CuFeS2 ==> 2000 x 64 / 184 Cu = 695.7 tonne copper
This is the maximum amount of copper that can be theoretically extracted from the 'pure' ore.
In reality there are impurities in the ore (e.g. other minerals) and in the extracted molten copper.
You can solve reacting mass problems (i.e. solve the ratio problem) with a series of logical steps set out in a table illustrated and explained below ...
Pick out the particular ratio you need to solve the reacting mass problem e.g. the particular reactant
and product
comments
CuFeS2 ==> Cu
only these bits of the equation are needed to solve the problem, you can ignore the rest of
the equation
184 tonne ==> 64 tonne
basic reacting mass ratio from the balanced symbol equation and the relevant atomic or
formula masses
then scale up by factor of 2000 for the initial 2000 tonne of ore
Therefore 695.6 tonne of copper can be extracted from 2000 tonne of chalcopyrite copper ore
o (b) If only 670.2 tonne of pure copper is finally obtained after further purification of the extracted copper by electrolysis, what is the % yield of the overall process?
o % yield = actual yield x 100/theoretical yield o % yield = 100 x 670.2 / 695.7 = 96.3%
o More on % yield and atom economy in calculations section 14.
o
- Reacting mass calculation Example 6a.6
o Calculating the theoretical yield of iron from an impure iron oxide ore.
o A sample of magnetite iron ore contains 76% of the iron oxide compound Fe3O4
and 24% of waste silicate minerals.
o (a) What is the maximum theoretical mass of iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon reduction? [ Atomic masses: Fe = 56, C = 12 and O = 16 ]
o The reduction equation is: Fe3O4 + 2C ==> 3Fe + 2CO2
o Before doing the reacting mass calculation, you need to do simple calculation to take into account the lack of purity of the ore.
o = 0.55 tonne Fe (550 kg)/tonne (1000 kg) of magnetite ore o
-o (b) What is the at-om ec-on-omy -of the carb-on reducti-on reacti-on?
o You can use some of the data from part (a).
o % atom economy = 00 x total mass of useful product / total mass of products
o Because of the law on conservation of mass, total mass reactants = total mass of reactants o % atom economy = 100 x total mass of useful product / total mass of products
o It doesn't matter which version you use for the atom economy calculation, you should get the same answer!
o = 100 x 168 / (232 + 2x12) = 100 x 168/256 = 65.6%
o More on % yield and atom economy in calculations section 14.
o
-o (c) Will the at-om ec-on-omy be smaller, the same, -or greater, if the reducti-on inv-olves carb-on monoxide (CO) rather than carbon (C)? explain?
o The atom economy will be smaller because CO is a bigger molecular/reactant mass than C and 4 molecules would be needed per 'molecule' of Fe3O4, so the mass of reactants is greater for the same product mass of iron (i.e. bottom line numerically bigger, so % smaller). This is bound to be so because the carbon in CO is already chemically bound to some oxygen and can't remove as much oxygen as carbon itself.
o Fe3O4 + 4CO ==> 3Fe + 4CO2
o so the atom economy = 100 x 168 / (232 + 4x28) = 48.8 % o Note reactants mass (232 + 4x28) = (3x56 + 4x44) products mass o More on % yield and atom economy in calculations section 14.
o
- Reacting mass calculation Example 6a.7 o Deposition from hard water samples
o On analysis, a sample of hard water was found to contain 0.056 mg of calcium hydrogen carbonate per cm3 (0.056 mg/ml).
o If the water is boiled, calcium hydrogencarbonate Ca(HCO3)2, decomposes to give a precipitate of calcium carbonate CaCO3, water and carbon dioxide.
o (a) Give the symbol equation of the decomposition complete with state symbols.
Ca(HCO3)2(aq) ==> CaCO3(s) + H2O(l) + CO2(g)
o (b) Calculate the mass of calcium carbonate in grammes deposited if 2 litres (2 dm3, 2000 cm3 or ml) is boiled in a kettle.
where z = unknown mass of calcium carbonate
z = 112 x 100/162 = 69.1 mg CaCO3
since 1g = 1000 mg, z = 69.1/1000 = 0.0691 g CaCO3, calcium carbonate
o (c) Comment on the result, its consequences and why is it often referred to as 'limescale'?
This precipitate of calcium carbonate will cause a white/grey deposit to be formed on the side of the kettle, especially on the heating element.
Although 0.0691 g doesn't seem much, it will build up appreciably after many cups of tea!
The precipitate is calcium carbonate, which occurs naturally as the rock limestone, which dissolved in rain water containing carbon dioxide, to give the calcium hydrogen carbonate in the first place.
Since the deposit of 'limestone' builds up in layers it is called 'limescale'.
Reacting mass calculation Example 6a.8o
-o This is a much m-ore elab-orate reacting mass calculati-on inv-olving s-oluti-on concentrations and extended ideas from the results.
o In this exemplar Q I've used the formulae a lot for short-hand.
o A solution of hydrochloric contained 7.3 g HCl/dm3.
o A solution of a metal hydroxide of formula MOH was prepared by dissolving 4.0g of MOH in 250 cm3 of water.
o M is an unknown metal but it is known that the ionic formula of the hydroxide is M+OH-. 25cm3 samples of the MOH solution were pipetted into a conical flask and titrated with the hydrochloric solution using a burette and a few drops of phenolphthalein indicator.
o All the MOH is neutralised as soon as the pink indicator colour disappears (i.e. the indicator becomes colourless).
o On average 19.7 cm3 of the HCl acid solution was required to completely neutralise 25.0 cm3 of the MOH solution.
o [Atomic masses: H = 1, Cl = 35.5, O = 16, M = ?]
(a) Give the equation for the reaction between the metal hydroxide and the hydrochloric acid.
MOH(aq) + HCl(aq) ==> MCl(aq) + H2O(l)
You may or may not be required to give the state symbols in (), or you may be just asked to complete the equation given part of it.
(b) Calculate the mass of HCl used in each titration.
1 dm3 = 1000 cm3, so in 19.7 cm3 of the HCl solution there will be
19.7 x 7.3 / 1000 = 0.1438 g HCl
(c) Calculate the mass of MOH that reacts with the mass of HCl calculated in (b).
25cm3 of the 250 cm3 MOH solution was used, so the mass of MOH titrated is
25 x 4 / 250 = 0.40 g MOH
(d) Calculate the formula mass of HCl.
formula mass = 1 + 35.5 = 36.5
(e) Calculate the mass in g of MOH that reacts with 36.5g of HCl and hence the formula mass of MOH.
0.1438 g HCl reacts with 0.40 g MOH