What are the mass and the identity of the precipitate that forms when 55.0 mL of 0.100 M BaCl2 reacts with 40.0 mL of 0.150 M Na2CO3?

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Hello. Everyone in this video. We need to go ahead and first identify and then next to find the mass of the precipitate that's formed when this reaction here occurs. So we go ahead and recognize the dissociation of our two materials here. So that's starting off with C. U. C. L. Two, which is that I'll go ahead and dissociate into its ionic form. So let's see Youtube plus And two moles of cl -. Our second material here is N. A. O. H. Which of course a chris now go ahead and associate into an A. Plus as well as O. H. Minus. So when we have these two materials reacting with each other, it's a double displacement reaction. So basically what that means is our kid ionic and an ionic partners are being switched. So what we'll have then is R. C. U. Two plus reacting with instead of cl minus O. B. R. O. H. And then go ahead and yield C. U. O. H. Two which is a solid. And the second one will be when R. N. A plus reacts with the cl minus which yields an A. C. L. And that's Equus. So we see then that because we have a solid forming, that's a precipitate and that's of course insoluble. And that's what our identity of our precipitate is. So let's go ahead and highlight that as our first answer for one or I guess i in our case the second is to go ahead and solve for the math of the precipitate that's formed. So let's first go ahead and write out the actual chemical reaction that occurs. So again, we have a certain materials of CUC. L two reacting with N A O H. That yields CUOH two, which was as solid as well as NACL. All right, So what we'll do now is to go ahead and bounce up. So let's go ahead and draw a line to separate our products from starting materials. So we have adam, C U C L N A O and H. Same thing for the other side, of course. So, I'm basically just rewriting that. And here we're just essentially keeping a tally of all the atoms that we have to make sure everything is balanced. So just starting off without any coefficients, we have one copper to chlorine, one sodium, one oxygen and one hydrogen on the product side, we have one copper, one chlorine, one sodium, two oxygen and two hydrogen. So what catches my eye first is the oxygen and hydrogen. So we want to go ahead and balance that. So we see that our source of oh, NH is from N A O H. So I ought to go efficient of two. This will go ahead and give me two moles of sodium, two of oxygen and two of hydrogen. And then for the right side, I'm gonna go ahead and balance out my sodium as well as my chlorine. So of course, which is a coefficient of two to N A C. L. So then we get two moles of chlorine and two moles of sodium. So this is going to be our balanced ration that we use for our dimensional analysis. So now, since we have two different products, we essentially need to find our lemony reagents and however much product that produces is how much we'll actually have. So we go ahead and use Our first start material of c. two were given 15 ml of this. We're gonna convert this into leaders and we use the concentration as a factor here. So we have 1000 ml for every one liter. Then we can go ahead and use polarity. So for every 0.200 moles of our story material, we'll have one leaders of solution. We see now that the milliliters cancel as well as leaders. Then we can go ahead and do a multiple ratio. So for our star material and our precipitate here, everything is a 1 to 11 to one ratio. So for every one mole of our star material We'll have one mole of our precipitate, Which is CUOH two. All right, so the moles will cancel. And next we have to go ahead and multiply this by the molar mass of our precipitate to get us the mass. So for every one mole of C U O H two R precipitate. We'll go ahead and have 97. g of our precipitate. So you can see that the moles of precipitate will cancel. All right? So now, once I put everything into my calculator, I'll get the value of 0. grams of our precipitate. So that's only if we're using the first starting material here. Let's go ahead and sell for our second starting material. All right. So here we're using N A O H. So we're given 50 million years this time of N a O H and again, we're converting the millions into leaders with a direct conversion and then we'll go ahead and utilize our concentration as a conversion factor. So for every 0.180 moles of NAOH, we'll have one leader of the solution and then I want multiple ratio. So for every one mole of our product or our precipitate, we have two moles of any age. So two moles of n a O H for one mole of precipitate. Okay, now we see that our milliliters council's leader, councils of Nuh councils. And lastly, we just need to end this off with our molar mass of our precipitate here. So for every one mole of c u o H two, We'll have 97.561 g of precipitate. Okay, so again, we can see then the moles of precipitate councils. And that gives us a value of 0.4390. And of course the units being grounds of our precipitate. So now comparing these two values, we can see that this one is of course less lesser value. That means our first our material, RCUCL two will be our limited regions. So then we can say that therefore the mass of our CCL or the mass of are precipitated. C u o h two is equal to 0.2927 g. So this is going to be our final answer for the second part of this question.

What are the mass and the identity of the precipitate that forms when 55.0 mL of 0.100 M BaCl2 reacts with 40.0 mL of 0.150 M Na2CO3?

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Key Concepts

Precipitation Reactions

Stoichiometry

Molarity and Volume Calculations