Brass is an approximately 4:1 alloy of copper and zinc, along wit...

Question

Brass is an approximately 4:1 alloy of copper and zinc, along with small amounts of tin, lead, and iron. The mass per-cents of copper and zinc can be determined by a procedure that begins with dissolving the brass in hot nitric acid. The resulting solution of Cu2+ and Zn2+ ions is then treated with aqueous ammonia to lower its acidity, followed by addi-tion of sodium thiocyanate (NaSCN) and sulfurous acid (H2SO3) to precipitate copper(I) thiocyanate (CuSCN). The solid CuSCN is collected, dissolved in aqueous acid, and treated with potassium iodate (KIO3) to give iodine, which is then titrated with aqueous sodium thiosulfate (Na2S2O3). The filtrate remaining after CuSCN has been removed is neutralized by addition of aqueous ammonia, and a solu-tion of diammonium hydrogen phosphate ((NH4)2HPO4) is added to yield a precipitate of zinc ammonium phosphate (ZnNH4PO4). Heating the precipitate to 900 °C converts it to zinc pyrophosphate (Zn2P2O7), which is weighed. The equations are 
(1) 
(2) 
(3) 
(4) 
(5) 
(a) Balance all equations.

Accepted Answer

Hello everyone. So in this video we just need balance every single one of these five different reactions. So let's go ahead and get started. So first I'm gonna go ahead to scroll down to give us a little more space of course I'll start off with reaction one here. So for action one the first step I'm gonna do is to go ahead and break it to separate half reactions. So that's going to be P. D. Going to the ion version. So a four plus charge. And this is the second will be N. 03 minus that is a chris and that goes to N. O. Gas. Alright so the first thing I'm gonna do is go ahead and balance out our oxygen's. We see from the first half reaction we don't have any oxygen atoms so we don't need to worry about that. But the second one does. So for a starting material we have three atoms of oxygen but for the right we only have one animal oxygen to add oxygen atoms. We can add H. 20. So because the right side needs to more oxygen's we'll add two moles of H. 20. Of course that it's in its liquid state. The second step is to balance out the higher regions and we can see from the right side because of our H. D. O. S. We added four atoms of hydrogen on the left side we have zero. So go ahead and add that with hydrogen ions so add four H plus of course that's Aquarius and then we can go ahead and balance out our charges. So I'll do this in a different step because we're running out of space here but we can see for the first step here are the first half reaction on the star material side. We have a neutral side on the right side, we have a four plus charge to make the right side neutral. We can go ahead and add a negative charge which in case would be a electron. So to bounce out this four plus we can go ahead and add for electrons. Now for the second half reaction here we can see on the left side, well we have four plus ones. So we have a four plus with a negative one. So that's a plus three overall charger here on the story material side on the right side here, it's completely neutral. So to neutralize this side, we go ahead and add, let's see. Just go ahead and actually rewrite this. So again we have palladium solid going to its catatonic form and to balance out the charges were adding for electrons. And then for the second half reaction here we are starting off with the for hydrogen ions with one mole of n 03 minus which gives the side a Positive three charge. We're gonna go ahead and add three electrons to balance off this charge. So it can be neutral. Just like the product side. All right. So we're gonna eventually want to combine these two half reactions here but we see that the electrons are not balanced. So, that's the fourth step then, to balance out the electrons. How we can do so is if we multiply the first reaction all by three And multiply the second half reaction all by four. Now we do. So we can get four moles of solid palladium that goes to four moles of the palladium karan with a four plus charge. And then for electrons would be 12 electrons. And then for our second half reaction, That's going to be 16 hydrogen ions With four moles of RN 03 -. Of course. Now we have 12 electrons on the star material site for the second half reaction. And then for our products, we have four moles of N. O guests As well as eight moles of H 20. And its liquid state. So now we're gonna go ahead and combine these two half reactions. We can see That the 12 electrons will cancel each other out. So, the remaining balanced reaction is going to be I see a mistake here, there's going to be three moles of this palladium with solid And I just erased the four hydrogen plus. Alright, so now we can go ahead and combine them. So we will be left with On the source material side three moles of ours solid palladium reacting with another mistake when I race this, Is that four times four is actually 16 here. Alright, now we can go ahead and combined. So we'll have 16 hydrogen ions, Four moles of R. N. 03 minus an ion. That's it for our sarah material site for product side we of course have And again, one small mistake that I see here Is that we're multiplying by three. So we'll get three palladium Canadians. All right, so now we have three palladium four plus ca downs which is acquis that reacts with four moles of R. N. O. Gas as well as eight moles of H 20. So then this is going to be our balanced reaction. Alright, so same exact process for the second reaction here, I'll just go ahead and do a different color. I'll scroll down as well because we need more space. Alright, so for a second reaction, I'll just go ahead and rewrite it. So, original equation that is given to us in the problem is the palladium four plus cat island is reacting with S. C. N. Negative As well as HS. and that yields P. D. S. C. N two which is a solid as well as H S. 04 minus again. That is this Alright, so again, first step is to go ahead and separate this big reaction into two half reactions. So one of them will go ahead and be palladium four plus plus our S C. N minus. And that goes ahead to yield Lady. Um S. C. N. Two are solid here and the next half reaction is S. H. S. 03 minus to yield H. S. 04 minus. Again that is so first step I'm gonna do is go ahead and bounce up my non hydrogen and oxygen is first. So you can see from the first half reaction here that on the right side we have two atoms of sulfur on the left side. Sorry material side we only have one, so we'll go ahead and add a coefficient of two here. Alright now that everything is balanced for the non hydrogen and non oxygen's we can go ahead and now bounce out our oxygen's. So we see from the first half reaction we don't have any oxygen atoms so we don't need to worry about that. But the second one does. So for the starting material side we have three items of auction but for the right we only have four. So how we add a oxygen atom to the starting material size by adding water. Right, So we can go ahead and add that. We'll add one mole of H20. Since we only need one atom of auction. Alright so it looks like it's time to maybe rewrite this so be less cluttered. So again we have our palladium four plus cat ion that reacts with two moles of s. c. n. And that yields our neutral molecule that's a precipitate here. And the 1st 2nd 1 we have added H 20. In its liquid state with our H. S. 03 minus To go ahead and yield HS. -. Alright. The third step we're going to take is balance out our hydrogen. So we see from our again first half reaction here, we don't have any hydrogen atoms so we don't need to worry about it for the first half reaction. But for the second one we do so from the start materials since we either HBO, we have two atoms of hydrogen from HBO and one from the an ion. So the total comes out to be three hydrogen atoms. But for our product side we only have one. So we'll go ahead and add hydrogen atoms to the product side And we just need to hydrogen so we can add two hydrogen ions. And last thing we need to do before we combine the half reactions to balance on our charges. So we can see from the first half reaction then the product side is neutral but the left side we have a four plus charge in a and two negative ones. So we need to go ahead and neutralize this by adding two electrons. As for our second half reaction here this side we have a negative one charge on the right side. Let's see we have a negative one and then we have two plus one here. So go ahead and add another two electrons to make this side a negative one charge as well. Now that everything is balanced, we can see our electrons are also balanced. We have two electrons on the first half reaction on the starting material side. And the second half reaction, we have two electrons on the product side. So we go ahead and combine these now. And of course our two electrons will go ahead and cancel if we do. So our combined and balanced reaction is going to be where our palladium caddy on reacts with two moles of R S C N and I am As well as one more of H and one mole of our HS 03 and I on and that let's go ahead and rewrite this more clearly. And that goes ahead to yields P D S C N. As well as HS 04 negative and tools are H plus Canadian and that's going to be our balanced reaction for the second one here. Alright, next one, let's go ahead and scroll down again. I have a second page here because I knew we would need it. Alright, so I'll just go ahead and rewrite the original given third reaction. That's when our palladium to Pluskat ion reacts with io three minus an eye on and that reacts to give P. D four plus Katie on and iodine which is a quiz. Alright, so again, first thing we're gonna do is go ahead and separate this big reaction to two half reactions. So that's going to be where our first starting material here yields our palladium four. Plus cut down our second half reaction is when our an ion from our star materials goes ahead to yield iodine in its liquid state. Alright, so first thing we're gonna do is go ahead and bounce out are not high regions and non oxygen's. So of course our first half action, we don't need to worry about that. But on the second one we do. So you see on the product side, we have two iodine on the left side here, we only have one. So go ahead and add a coefficient of two here. Now we can go ahead and balance out our oxygen's. So we only need a concern that for our second half reaction here. So you can see from the left side that we have. Well two times three is six. We have six oxygen atoms here. On the right side, we have none. So we'll go ahead and add six moles of H 20. Now that our oxygen's are balanced. We can go ahead and bounce out our high jeans again, that doesn't concern our first half reaction here, but only our second. So I see here on the right side we have well six times two is 12. We have 12 high regions but on the left side we have none. So go ahead and add 12 moles of our hygienic cat ions here, of course that's in its liquid state. All right. And now that hydrogen and oxygen in our balanced we can go ahead and balance our charges. So we see here from the first reaction on the right side we have a four plus charge on the left side, we have a two plus two. Make this side equal to this. We can go ahead and add two electrons. Now they both have a two plus charge. Now for the second half reaction here. So for the right side is completely neutral on the left side we have positive charges and two negative. So that's 10 positive charges in total plus 10 charge. So to neutralize this, we'll go ahead and add electrons. And I've completely neutralized the starting material side. Now. All right, we can see that. Now we need to go ahead and balance out our electrons. So for our second half reaction, we have 10 on the first half reaction we have to so we just need to multiply the first half reaction by five. And then we do So let's go ahead and run out that reaction. So we'll get five moles of our palladium, two plus Karen, Which yields five moles of our palladium, four plus cast iron and 10 electrons for a second have reaction here, we'll have 10 electrons reacting with our 12 h plus cantons and two moles of r I 03 and ions. And that yields are iodine which is a glorious and six moles of H20 in its liquid state. So we go ahead and combine these two have reactions now, you see that are 10 electrons will cancel. Which leaves us with just five palladium cat ions are two plus ions that reacts with 12 moles of our ha jin carry ons and two moles of r I 03 minus which yields five moles of plating four plus Our iodine in its liquid state and six moles of H 12. So this is going to be our balanced third reaction here in the problem. Alright, moving on, we'll do our 4th. Go ahead again, scroll down. So for our fourth this one is a little bit more simpler. Go ahead again. Right out the given problem. So that's iodine and it's a quick estate reacting with S 2032 minus. Again, that is and that yields are i minus an eye on And S two minus which is. So again, we have to go ahead and separate them into two half reactions. The first one will be I too. Going to i minus and the second one will be S 2032 minus Giving S 406 to -. All right. So we can go ahead and balance out our non high regions and non auctions first. So for these first half reaction we can see on the material side we have to I dines on the right, we only have one. So let's go ahead and add a coefficient of two. And then for our second half reaction here, we can see that we have starting material side we have two atoms of sulfur on the right side, we have four, so get ahead and add a coefficient of two. Now we can go ahead and bounce out our charges since everything else is balanced. So for our first Half reaction here, we see on the side of material side is completely into neutral. On the right, we have a -2 charge to neutralize the siding material side. We'll go ahead and add two electrons. And then for the second half reaction. Then on the left side here we see we have two times two is four. So we have a negative four charge on the right side, we have negative two. So to make this more negative, we'll add two more electrons. Alright, so the electrons are bound so we can go ahead and just combine our half reactions and we can see that of course our electrons will cancel out here. And that just leaves us with one mole of I two That reacts with two moles of S203, which goes ahead and yields to mole of i minus As well as S 406 to -. That's going to be my balanced fourth reaction of this problem, we just have one more left again. I'm gonna scroll down a little bit. All right, So for our final one I'll go ahead and again rewrite our given equation. So we have C A N H four P. 04 which is a solid, which yield C A two P 207, A solid as well as water in its gasses form and lastly and age three and its gasses form. All right, so, how we can go ahead and balance This is different from how we've done in the first four. So, this is just the regular way of balancing out an equation. So, what I'm gonna do is first draw a line here that separates our star materials from our products. And then I'll go ahead and keep a tally kind of of all the different items that we have. So, of course we have C A N H. P and O. And that's going to of course be the same for our other side. So, it could just be right what we wrote so far, H. P and O. All right, so, just starting off without doing any sort of balancing here, we see we have one mole of calcium, one mole of nitrogen, four moles of hydrogen, one mole of our phosphorus and four moles of our oxygen on them. For the right side. We have two calcium, one nitrogen five hydrogen to phosphorus and eight oxygen's. Alright, so, you see here that I wanna go ahead actually, first balance out my calcium. So we add a coefficient of two in front of our only starting material that goes ahead to give us then to calcium is to nitrogen, eight hydrogen ins to phosphorus and eight oxygen's and how it balances on the right side. Then we need to go ahead and balance out our nitrogen. So I was at a coefficient of two. If we do so then we get to nitrogen and total of eight high regions, so we see now that everything seems to be balanced here. So then this right here is going to be my Balanced reaction for our 5th equation and this problem.

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