Dilution of "Stock Solutions"Assigned Final Volume Molarity of "Stock Solution"Z50nL LOml Sel [soln Volume of "Stock Solution 025 L zn Mola...

$$ \begin{aligned} &\text { Calculate the molarity of } \mathrm{Zn}^{2+} \text { ion in a saturated solution }\\ &\text { of } \mathrm{ZnCO}_{3} \text { , that contains } 0.25-\mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} \text { . } \end{aligned} $$

All right. So if we want to calculate mole aren? T we need to recall the fact that more clarity symbolised by capital M. Is equal to moles per liter. Right? Specifically moles of solute the thing that's getting dissolved over leaders of solution. So we're gonna need moles are gonna need leaders to be able to solve this. Now. We were given a chemical equation. We have some zinc and some copper chloride which is gonna forms inc chloride And copper. And if if we have 25.0 g of zinc, how much zinc chloride can we make? So we're going to convert first the grams of zinc into moles of zinc chloride. That will be our numerator for polarity. And if we know the volume of the solution, we can convert that into leaders and then divide the two numbers. All right. So let's do a little bit of strike geometry. We have 25.0 g of zinc. We can use molar mass to convert into moles. And uh the molar masses Incas 65.39 g per mole. And then we want to go to zinc chloride. So we use the mole ratio coefficient for zinc chlorides. One coefficient for zinc is one. So for every mole of zinc chloride we make, we have to use one mole of zinc and we can see that grams of zinc and mold of zinc cancel out. We're left with moles of what we want Our is in chloride and that is .382 moles of our salute zinc chloride. Now we have our moles. So we have the value that we're gonna plug in. Um For our moles. Leaders though, we're told that we have 275 ml of solution, we have to convert that to leaders. And a leader is the same as 1000 ml. So that is .275 leaders. I've got moles, I got leaders, so I can now solve for More clarity, and I'm going to get a value of one 39 moles per liter To three significant figures.

This question contains multiple parts and requires you to review the concept of limiting reactant. We have to react INTs, both of which we are, for both of which we are given quantities. Both of these quantities will decrease, one will decrease to zero. That will be the limiting reactant. The other one will decrease not to zero, and some will be left over. The maximum amount of product that can be made will be produced when the limiting reactant is completely consumed. To identify the limiting reactant, you need to calculate the amount of product that can be created by each reactant, and the one that produces the smallest amount of product will be the limiting reactant. So part A asks you to calculate the moles of precipitate that will form in the reaction. To do this, you will calculate the moles of precipitate that will form for each of the reactant. If each were consumed completely and the one that produces the smallest amount will be the limiting reactant. Before we can do that, we must first identify the chemical reaction that is occurring. We have zinc nitrate to reacting with potassium hydroxide. If zinc nitrate reacts with potassium hydroxide and the cat eye on switch places will get potassium nitrate, which, according to the scalability rules, is soluble and zinc hydroxide, which, according to the scalability rules, is insoluble. So to simplify things, the Net Ionic equation is zinc two plus reacting with two hydroxide producing the insoluble precipitate zinc hydroxide. Everything else are spectator ions. Let's now calculate the mass of the precipitate zinc hydroxide that can be created if all of the zinc nitrate were consumed. All 20 mil 200 mL will convert the 200 mL into leaders and then convert those leaders into moles of zinc two plus or zinc nitrate. Because the stock er the relationship is 1 to 1 by multiplying by the molar mass when we know the moles of zinc two plus, we can use the strict geometry up here to calculate the moles of zinc hydroxide, which can be created. It's a 1 to 1 relationship also, and we get 0.0 to 00 moles, zinc hydroxide if all of the zinc nitrate were consumed. Now let's figure out the moles of zinc hydroxide that can be produced from the potassium hydroxide, the 100 mL of 1000.1 Moller potassium hydroxide will convert the mill leaders to leaders and then use the polarity to get moles of potassium hydroxide, which is also equal to moles hydroxide. Once we know the moles hydroxide, we can use the stock geometry of 2 to 1 to calculate the moles of zinc hydroxide that can be created from this many moles of hydroxide being consumed. And we get 0.500 moles of zinc hydroxide. So because the but the potassium hydroxide produces the smallest amount of zinc hydroxide, potassium hydroxide is the limiting reactant. The hydroxide then is consumed completely, and we have some zinc nitrate or zinc left over. The amount of zinc hydroxide precipitate that is created would just be the 0.5 moles. Now, for the second part, let's calculate the concentration of the zinc ions that are left in the final solution. To do this, we need to know the moles of zinc that did not react, and we need to know the total volume if we have the moles of zinc that did not react and we divided by the total volume of solution in leaders. That ratio is the more clarity of zinc ion left in solution. So let's start with the moles of zinc that did not react. To do this will take the volume of zinc nitrate 200 mL or 2000.2 leaders multiplied by its concentration to get the moles of zinc that we start with. Then we can take the 100 mL of potassium hydroxide converted to leaders 0.1. Leaders multiply it by the concentration of potassium hydroxide to get moles, potassium hydroxide. But we know from the stock geometry that one mole of zinc is consumed for every two moles of hydroxide that are consumed. So this is now the moles of zinc that are consumed. This is the molds of zinc we started with. This difference is the moles of zinc that are left over. We then divide that by the new volume, which is the 200 mL plus the 100 mL that were mixed together. And we get a zinc polarity left in solution a 0.50 Moller

Have you even saw in this question Calgary the modularity of solution that contains a two intertwined dash minus three mol. Global seeking self at in 35 ml of solution. Be to 350.75 more operation. Hydroxide in total volume of 1. 35 ml of solution. So a part H two A cell phone. Yes to edge plus plus it's a fall to minus similarity of their support. Okay. And still suffer. That is equal to point 17. More. Yeah. Extra so far upon 85. M L wow. A spirit so full. Mhm. Mhm. Mhm. Into 1000 ml Extra so far divided by one later. H two A cell phone. Mhm. Mhm. Mhm. And therefore. Mhm More clarity. Mhm of a stress cell phone is equality. Two more looks. Mhm. Then bees. Yeah. A man. All three awesome. Yes. N A plus plus. I know three baroness. So modularity of a man would be that is equal to 7.5 in 2 10 national minus three mold. Oh and make another three. Yeah. Okay. Divided by 13 points. Fire. Matloff. Any and all three into 1000 men. Yeah. A man or three. Mhm. Divided by one leader and then on three. Yeah. And therefore malaria. T. Of an N. Or three. Oh, yeah. Yeah. That is equal to $0.56. Uh huh. Yeah.

How would you answer in this question? There's calculate majority of solution that contains a to international minus three more. COBOL celebrating 35 Ml solution. So we'll begin with a part. Come on. Self ed. Yeah. What? Yes zero to plus plus I saw four to minus modularity that is equal to and upon. Well that is equal to two in 2, 10 to minus three mold cobalt sulfide upon 35 miles. Mhm koval self fed into 1000 and men of global self raid, a bug, one liter cobalt sulfate and therefore more clarity. That is equal to 0.0 57 molars. Yeah. Who developed? Okay wage gives K plus plus h minus similarity of garage of coverage that is equal to and upon. We that is equal to 2.75 mol carriage upon 1 35 AM allowed to wage into 1000 ml carriage upon one liter cage and then therefore McGarity of kuwait that is equal to 20.37 moller. Thanks love