Name: Slide – Exercise 15 - Metals
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Analyse the process of extracting copper to account for the imperfectly yield percentage

Analyze the process of extracting copper to account for the imperfect yield percentage, which is inevitable for all extraction processes.
OK. So this question is asking a lot of it. But all it’s saying is look at the process of extracting copper and account for why we don’t get perfect copper yield percentages.
Why don’t we get 100 percent copper out when we try to use this extraction process?
Well, the yield of copper is not 100 percent because – there’s a loss of copper during milling as not all copper is skimmed off the top. So during the froth flotation step, when we skim the copper off the top of the water – so if you have no idea what I’m talking about, just look back on the copper extraction video and you will understand.
But when you skim the copper off the top, you may lose some of it or you may not get all of it. So that’s one part that adds to this 100 percent yield that we can’t get.
Smelting processes do not cause all the copper to react. Some reactants or products are lost via vaporization due to extremely high temperatures. So even though it’s a metal, the extremely high temperatures mean we do get some metal vapours and so we do lose some copper when we do the smelting process, so when we heat it in the oxygen, to remove the sulfur. Again if you don’t understand, look back on the copper video.
So we lose some via vaporization and not all of the copper will react simply because the surface area may not be perfectly tuned for that or the surface may have corroded a little bit and it’s protecting the inner layers.
So we need to find a way to deal with that. OK? So the smelting process isn’t 100 percent either.
Copper one oxide then reacts with the unreacted copper sulfide to produce pure copper with sulfur dioxide. Eventually, there will be not enough unreacted material to continue the loop. So eventually we run out of unreactive copper.
The reaction may be too slow and completion may be economically not viable. So some reactions don’t happen fast enough. So we may not bother and so we lose some of the copper simply because it’s too hard to get that last percentage out.
Side reactions where reactants form secondary unwanted products. This happens a lot actually, this one. In lots of different reactions, you get side products, which actually stop us from getting what we actually want and that – and all of these contribute a small amount to the process. They all contribute a small amount and so that’s why we can never quite get 100 percent. OK? So that’s …

Chemical Equations that Crystal Zinc Dissolved in Dilute Sulfuric Acid

A crystal of zinc weighing 31.4 grams was dissolved in dilute sulfuric acid forming zinc sulfate and hydrogen gas, which was collected. The gas was dried and then completely burned to give 8.67 grams of water.
Write chemical equation for these reactions. Now this is a hard calculation. It looks like it’s fairly straightforward. But there’s a lot to this question. The reason why it’s hard is because it’s multi-step. This is one of those ones where it’s good to pay attention now because this is something that if you learn how to do now, you will have no problem in titration, which is considered one of the more painful topics in chemistry.
Chemical equations. The first one is zinc dissolves within the H2SO4 to release hydrogen and zinc sulfate. A simple equation.
Then the hydrogen gas is burned with oxygen to produce H2O.
So deduce the mass of the zinc crystal. So we had a lump of zinc. But we don’t actually know how much of it was actually zinc. So we want to know the mass of the zinc inside that crystal.
So we start with this reaction and this reaction. So the mass of the water is the only mass that we’re given. So we calculate the number of moles of that. So 8.67 divided by the molar mass of water, which is this. It gives you 0.4795 moles. OK?
Now from the second chemical equation, the number of moles of water equals the number of moles of hydrogen. So the number of moles of hydrogen is also 0.4795 moles.
Then you might be wondering, “Well, why in the heck would I bother wanting to know how much hydrogen comes out?” Well, if you look at that equation, what links the two equations? The amount of hydrogen because the hydrogen that comes out of here is burned here. So the hydrogen that’s consumed here must be the same as the hydrogen from here.
That’s why we want to have the number of moles of hydrogen there. OK? From the first chemical equation then, the number of moles of hydrogen equals the number of moles of zinc, which is again 0.4795.
So the mass of the zinc is just the moles of the zinc times the molar mass, which is 0.4795 times 65.38, which is 31.35 grams. OK? So you can see that throughout this question, we’ve been – the reason why it’s difficult is because there are all these linking parts that you have to still work your way through.
If we told you all the steps, then that would be easy. But the way to do this question is just to work back and calculate things that you know until – and then link things that are supposed to be linked and then you get your answer.
So that’s this question. It’s a very, very typical style of question that they ask, lots of link to parts and so it’s just up to you to sort of filter your way through all the information.