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Nickel(Il) chloride reacts with 0.479 of" sodium phosphate precipitate nickel (II) phosphate and 0.958 of nickel(Il) phosphate are produced.Write the balanced ...

Question

Nickel(Il) chloride reacts with 0.479 of" sodium phosphate precipitate nickel (II) phosphate and 0.958 of nickel(Il) phosphate are produced.Write the balanced chemical equation for the reaction.How many EIS of Niekel(II) Chloride sodium phosphate? are needed react with 0.479Calculate theoretical yield of this reaction.Calculale = the percent yield of this reaction,

Nickel(Il) chloride reacts with 0.479 of" sodium phosphate precipitate nickel (II) phosphate and 0.958 of nickel(Il) phosphate are produced. Write the balanced chemical equation for the reaction. How many EIS of Niekel(II) Chloride sodium phosphate? are needed react with 0.479 Calculate theoretical yield of this reaction. Calculale = the percent yield of this reaction,



Answers

Chlorine forms from the reaction of hydrochloric acid with manganese(IV) oxide. The balanced equation is:
\begin{equation}
\mathrm{MnO}_{2}+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}
\end{equation}
Calculate the theoretical yield and the percent yield of
chlorine if 86.0 $\mathrm{g}$ of $\mathrm{MnO}_{2}$ and 50.0 $\mathrm{g}$ of $\mathrm{HCl}$ react. The
actual yield of $\mathrm{Cl}_{2}$ is 20.0 $\mathrm{g}$ .

Mhm problem number 85 is a percent yield problem in which we are given the experimental or the actual yield. But we will need to calculate the theoretical yield. So the actual yield, the experimental yield. What they actually obtained was 1.262 g Of the S two Cl 2. All right. So, we need to use the balanced equation given here in the problem. And we need to calculate the theoretical yield. What we should act, what we should produce based upon strike geometry or mathematical calculations, assuming we start with The five g of SCL two that were given. So I need to convert from five g of SCL two on the reactant side to the number of grams of S two cl two produced on the product side. So, this is a master master of geometry to find that theoretical yield. The first thing I need to do is divide by the molar mass of SCL too And that is 102.97 g of SCL too in every mole. So grams of SCL two will cancel. Next. I need them to use the mole ratio to convert from moles of SCL 22 moles of S two cl two. So looking at that balanced equation, I need those coefficients in front of SCL two is a three. So that means for every three moles of SCL too. Looking at the product side, there is no coefficient in front of S two cl two. So it's understood it's a one. So we can make one mole of us too cl two. Now I need to convert moles of S two cl 2 into grams by using its smaller mess And it's smaller masses. 135 0.4 grams of the S two cl two in every month. Okay. Yeah. All right. So this will give us grams of the product that we should have made are theoretical And I get to 186 grams of S two cl two. And again, this is our theoretical this is how much math tells us we should have made. The real world is very different. Our yield is often times less than the theoretical yield. As is the case here. The experimental yield was only 1.262. So to calculate percent yield. Mhm we need to take that experimental yield what they actually got when I did the experiment. And we need to divide it by the theoretical yield. Yeah, We will change that decimal 2% by multiplying it by 100, adding that percent sign. And we'll round three significant figures since 5.00 had three. So we will get 57 0.7% as our yield. So just barely better than half of what we should have gotten. Right And that's our final answer. Thanks so much for watching. Okay

This problem requires a stoy geometric relationship between a reacting reactant and a product that is formed. The psychometric relationships between these two species are found as the coefficient of a balanced chemical reaction. The balanced chemical reaction that we are considering is provided in the problem. It's three moles of nickel to chloride. Aquarius Reacting with two moles sodium phosphate, Aquarius producing one mole of nickel to phosphate, solid, and six moles of sodium chloride. So, to determine the mass, I'm sorry, the moles of nickel to chloride required to produce .517 moles of nickel to phosphate. We'll start with the molds, nickel to phosphate. Recognize that we need three moles of nickel to chloride to produce one mole of nickel to phosphate. So the relationship is 3 to 1, Giving us 1.5 1.55 moles of nickel to chloride required to make .517 moles of nickel to phosphate.

So now we're gonna work on problem 37 from Chapter eight. In this problem were asked about a mixture the reaction between potassium chlorate and barium nitrate, which is a precipitation reaction to for Miriam Chloride. So we find that we have a massive 2.45 grams and they ask us to determine women in reacting theoretically yield hand her sent you. So First, let's go ahead and put our reaction which is to potassium chlorides quiz plus barium nitrate, which is also our quiz. And this is producing barium chloride, which is the solid in question and calcium nitrate. Sorry, potassium nitrate. So we're given volumes in concentrations for both, uh, potassium and potassium chlorate and barium nitrate. So we need to multiply those together to get molds. And the reason we need to do this is to determine which is the limiting reactive. So the mat moles of K. C. L is 1.20 similarity times volume, which is you're a 0.250 leaders or 25 milliliters, and we get a value of 0.30 Mrs. Then four Mel's of burying nitrate. We have the concentration of 0.900 polarity and remote by 15 millilitres, or 0.150 leaders and our moles here 0.0 135 moles. Now we can't. In most cases, we can't simply look at the one that smallest to determine the limited re agent. We need to look at the ratio. So for every, uh, we need to potassium chloride for everyone, very matri. So we have, let's say, 30 moles of calcium chloride or 30 Miller and completely react with that. We would need, uh, 15 moles of barium nitrate, which we don't have. So this is the limiting re agent. It works out that it's the one that smaller here, but it will not always do so. So you always need to take into account the Starkey on a tree. So this is the limiting re agent. Now we can catch await the theoretical yield, which is just based off the starting amounts and historic geometry. So since 0.135 is our, uh, the menu every agent, we're gonna start with this. So we have 010135 moles of barium nitrate and there's a 1 to 1 ratio between barium chloride in barium nitrate, and then we need to multiply by the molar mass of very um chloride. So we get to, 8.23 grams for one more, and this gives us a value equal to 2.811 grands. So this is our theoretical yield. Now, the question is what is our percent healed. So the calculate percent yield all we have to do is, uh, put our actual which we're told in the problem is 2.45 grams over our theoretical 2.811 and then multiply by 100. And when we do this, we get a value of 87.2%. So we recovered or re able to produce 87% of the mass that we should have

For problem. # 82. This is a mole ratio problem. And for any mole ratio problem, we need a balanced equation. So, I want to start off by copying the balanced equation from the problem. So, we have nickel three chloride or nickel to chloride. Sorry, it's reacting with sodium phosphate and it produces nickel to phosphate and sodium clark. Okay. And I left the states of matter out because those are not important for this problem. So, for number 82, We are trying to produce .715 moles of the nickel to phosphate. And the question is asking us to determine how many moles of the nickel to chloride. We would need to do that, assuming there is plenty of the other reactant. All right. So, what we need to do here is use our mole ratio and the mole ratio comes from the coefficients in the balanced equation. So, it tells us that for every three moles of the nickel to chloride, I can make one mole of the nickel to phosphate. Since there is no coefficient written here, it's understood that it's a one. Right? So, that means I just need to do one step here. In my calculation, I need to use that mole ratio as a conversion factor. So, starting with .715 moles of the nickel to phosphate. Yeah. Yeah. Yeah. Mhm. And then using the mole ratio as my conversion factor. So, for every one mole of the NI three P 042. Yeah. Okay. I need to start with three moles. Yeah. Of the nickel chloride. Mhm. All right. And that's the only step I need to do because moles of the nickel to phosphate will cancel. And when I calculate this, that will give me the number of moles of nickel to chloride that we need. So calculating this. And rounding to three significant figures since .715 that we started with had three. Gives me 2.15 moles, Yeah. Of the nickel to chloride. And that would be our answer. Yeah. Thank you so much for watching.


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