5

STOICHIOMETRY=Solving limiting reactant problems in solutionSuppose 2.46 g of barium acetate dissolved in 350. mL of 31.0 m Maqueous solution of sodium chromate. Ca...

Question

STOICHIOMETRY=Solving limiting reactant problems in solutionSuppose 2.46 g of barium acetate dissolved in 350. mL of 31.0 m Maqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution_ You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it.Be sure your answer has the correct number of significant digits_Dxo

STOICHIOMETRY = Solving limiting reactant problems in solution Suppose 2.46 g of barium acetate dissolved in 350. mL of 31.0 m Maqueous solution of sodium chromate. Calculate the final molarity of barium cation in the solution_ You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits_ Dxo



Answers

A solution containing 1.63 grams of barium chloride is added to a solution containing 2.40 grams of sodium chromate (chromate ion, $\mathrm{CrO}_{4}^{2-}$ ). Find the number of grams of barium chromate that can precipitate. Also determine which reactant was in excess, as well as the number of grams over the amount required by the limiting reactant.

Let's first start by reading the balanced chemical reaction for barium chloride and sodium chrome eight, which will produce berry um, chromite and sodium chloride equals. Take this, Sir Precipitate equals we needed to hear to balance. We're told that we have 1.63 g of barium chloride and 2.40 g of sodium crow mates. So our first question is, find the mass Iberian chromite that can precipitate. So we're going to calculate the mass Siberian crummy from both starting reactant. So starting from 1.63 g of barium chloride and the massive Muller, massive barium chloride is 208.24 grounds in one more reaction stroking commentary from our balance equation. One mole of barium chloride that's equivalent to one more of barium crow mates and one mole of barium crow mates has a molar mass of to 53.34 grams of barium chrome it and working this out, we find that we were produced 1.98 g of barium crewmates from 1.63 g of barium chloride. Let's go to our second reactions. We have 2.40 g of sodium chrome. Eight. Convert this two moles. Solar masses 1 61 98 grams of sodium crow mates in one more reaction. Strike geometry one more sodium crow mates toe one more vary, Um, Crow mates and Mohler, Mass. Here to 53.34 grams of barium crow mates present in one mole of Barium Crow mates, and this would be to 3.75 grounds of barium crow mates. So we're going to compare these two numbers here, and we can see that the lesser amount of Aaron Crow mate is produced here So we can say that 1.98 g of berry, um, chrome eight will be produced. It's always the lesser amounts, because the barium chloride is are limiting reactant, and our sodium crow made would be our access. Reactant. No. For a second part of the question, we had to figure out how much of the excess reactant is required by the uhh mass in grams of the excess in excess of the amount required by the limiting reactor. So we're going to start with our limiting reactant 1.63 g of barium chloride, and we'll find out how much sodium chromite needs to react with it, so to 8 to 4 g vary. Um, chloride in one mole. Reckon Sochi? Um, a tree. One more library, um, chloride toe, one mole of sodium crow mate and one more of sodium crow mates has a Mueller Mass. Of 1 61.98 g of sodium crow mates, So this will give us 1.27 grounds of sodium crow, mate. So this is the amounts required to react with 1.63 g of barium chloride for a complete reaction. And the mass of the sodium crow mates uh, left over in excess un reacted would be the initial amounts 2.40 g, minus the 1.27 g. We find out that the mass left over in excess is 1.13 g of the sodium chromite

The solution is known as the solvent and the minor component or components are called the solid. So many chemical reactions are carried out in solutions and solutions are closely related to our everyday lives. So for the air we breathe, the liquids we drink and the fluids in our bodies, they are all considered solutions. So we have been carrying out a brief calculation. So what we have is the following the mass percent of the soviet that is equal to the mass of the solid divided by the mass of the solution. Multiplied by 100. We can then re arrange that for the mass of the solid where the solids, barium chloride, we can have the max. So you can calculate the mass half barium chloride. That's equal to 4 to 7 grounds. Well supplied by 3.17% divided by 100%. What we get is 13.54 g. Because our units of percent. Council will effort grams as expected. So then we can calculate the number of moles of barium chloride. So we have the following equation. So what we have is which is equal to 13.54 g, multiplied by one mole b a c l, divided by 28.23 grounds, b a c o two. So we get a 6.50 times 10 to the minus two, so that corresponds to see.

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

It's in 44. A genetic reaction. A plus three b. Okay. Giving scene you are The limiting reactant is to be found in four different cases. Yeah, In one more eat four more people. We live one more area. We're only three more Be so one more. We will be remaining after the reaction. So he will be doing everything reacted. He will be totally consumed. Limiting reactive. Well in the second case, two walls of a and three months off. Well, three moles off. We will require only one more off him. As for the equation, so one more off the these access. So we eat a limiting reactions In the third case on five or eight on 1.6 more V 1.6 more lead. Now the ratio between A and B is one is 23 So if you multiply 0.5 by three we get one point White what? 1.6 more off these present. So find one more off is access. Want one more access? What he will be completely consumed. So a is elevating Lee happen And the fourth Gale's 24 moles off and 75 modes of B R Yeah. And we can see the amount of being what is three times that of a So 24. Entertain is 72 with 72. So only 72 is You've got to completely lived with me 24 g of 24 miles off cape. But there there is. There are 75 moles. That means three more chances. Is there anything reacted?


Similar Solved Questions

5 answers
The rte constant tor thls zero-order reacicn 0.0270 M:at 30OproductsHow long (in seconds) would Iake ior tht concenirator 0l 4Der*ase JIrnm900 M 0.240 M?NumborToole
The rte constant tor thls zero-order reacicn 0.0270 M: at 30O products How long (in seconds) would Iake ior tht concenirator 0l 4 Der*ase JIrnm 900 M 0.240 M? Numbor Toole...
5 answers
Problem 11.5PanA(Linr ] KhmiFullt UlFerOne umplEic 7 8 Ns? Ernice& VOUantter signllicant Inutte and iucluueannromialieValuteUnitsSebeminBxmka Eeri dGontinue
Problem 11.5 PanA (Linr ] KhmiFullt UlFerOne umplEic 7 8 Ns? Ernice& VOUantter signllicant Inutte and iucluue annromialie Valute Units Sebemin Bxmka Eeri d Gontinue...
5 answers
How can the following synthesis be carried out? COOHNO;
How can the following synthesis be carried out? COOH NO;...
5 answers
This table gives select values of the continuous and decrcasing function9(2)We'rc Intcrested in the area under thc curvo between 8 and $ 12,and we re considering uslng left and right Riemann sums, @ach wlth four cqual subdlvisions; t0 approxlmate ILOrder the areas from Ieast (on top) to greatest (on bottom)Left Riemann sumRlght Riemann LActual area under the curve
This table gives select values of the continuous and decrcasing function 9(2) We'rc Intcrested in the area under thc curvo between 8 and $ 12,and we re considering uslng left and right Riemann sums, @ach wlth four cqual subdlvisions; t0 approxlmate IL Order the areas from Ieast (on top) to grea...
5 answers
HF H Aud solid- the hcight of density of 0.540 gcm' with the lithium ) 1 slab cneity ab % inchcn ofmny 23 and hay 1 H by ILI advanced batterics: Wnnn slabdepicted Practice H Problem uon L8 L) Gingh unknown metal is added t0 - Density thed Oan S 1 'JssitieL, S PmL, iron) H and the- final volume-
HF H Aud solid- the hcight of density of 0.540 gcm' with the lithium ) 1 slab cneity ab % inchcn ofmny 23 and hay 1 H by ILI advanced batterics: Wnnn slab depicted Practice H Problem uon L8 L) Gingh unknown metal is added t0 - Density thed Oan S 1 'JssitieL, S PmL, iron) H and the- fina...
4 answers
10. Use Ar = 1 and At = { to numerically solve the heat equation "tI = Wtfor 0 <r<4,0 <1<1with boundary conditions "(0,t) = 0 = "(4.t) , aud initial condlitionu(z,0) =30(4- #)" for 0 < r < 4As a check. the bottom row shoukl Su (0 6 10
10. Use Ar = 1 and At = { to numerically solve the heat equation "tI = Wtfor 0 <r<4,0 <1<1 with boundary conditions "(0,t) = 0 = "(4.t) , aud initial condlition u(z,0) =30(4- #)" for 0 < r < 4 As a check. the bottom row shoukl Su (0 6 10...
5 answers
34. The first three terms in the expansion of (l + ayy" and 68y2 are 1, 12y, Evaluate a and n. Use the fact that (l+ay)" =l+nay+ In-l(ay)? (6 marks)35. Find the term not involving yin the expansion of (4 marks)
34. The first three terms in the expansion of (l + ayy" and 68y2 are 1, 12y, Evaluate a and n. Use the fact that (l+ay)" =l+nay+ In-l(ay)? (6 marks) 35. Find the term not involving yin the expansion of (4 marks)...
5 answers
ClelennemecmdInacrucoIn the hbontory you Are given the task of scpanting Ba"Iqucous solutlon Fot Gich reagent listed bclow indicate ifit can ba used scparate the !UrL$ Type [pt, pivc the fonnula of tbe precipitale . [f It cannol, tytc "No"for yes "N" for 1o, If te reagent CAN be uscdt Eepnnicta;Tor E Re#pcnt |Formula Prccipitate If YESNulK,soNaOHCul
Clelennemecmd Inacruco In the hbontory you Are given the task of scpanting Ba" Iqucous solutlon Fot Gich reagent listed bclow indicate ifit can ba used scparate the !UrL$ Type [pt, pivc the fonnula of tbe precipitale . [f It cannol, tytc "No" for yes "N" for 1o, If te reage...
5 answers
Point) tank contains IOOOL of pure water Brine that contains O4kE of salt per Iiter enters the tank at rate of SL/min. Also, brine that contains O9kg salt per Iiter enters the tank at a rate of 1OL /min. The solution kept thoroughly mixed and drains from the Iank : rale 15L/min. Answer the following questionsHow much salt is in the tank after minutes? Answer (in kilograms): S(t)How much salt is in the tank after hours? Answer (in kilograms):
point) tank contains IOOOL of pure water Brine that contains O4kE of salt per Iiter enters the tank at rate of SL/min. Also, brine that contains O9kg salt per Iiter enters the tank at a rate of 1OL /min. The solution kept thoroughly mixed and drains from the Iank : rale 15L/min. Answer the following...
5 answers
Consider the pentapeptide: Ala-Cys-Asp-Gly-Phe. A. At pH 4.5, what will the charges be on the two forms of thepeptide that will predominate?B. What will the ratio two forms be?
Consider the pentapeptide: Ala-Cys-Asp-Gly-Phe. A. At pH 4.5, what will the charges be on the two forms of the peptide that will predominate? B. What will the ratio two forms be?...
5 answers
Napel2] DETAILSPREMOUS ANSWIRSOscolphys2016 12.1,P,015. Hnmnnt [a ut 40,Jo9My ROTESAKtour TuAChcRPaacncAnea tha (owInor'h0t tha ccnelrutto WalteaUetnan~(un @n 0h Kel (4tn
napel2] DETAILS PREMOUS ANSWIRS Oscolphys2016 12.1,P,015. Hnmnnt [a ut 40,Jo9 My ROTES AKtour TuAChcR Paacnc Anea tha (ow Inor'h 0t tha ccnelrutto Waltea Uetnan ~(un @n 0h Kel (4tn...
5 answers
Boat is pulled into dock by pe attached to the bow of the boat and passing through pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at rate of m}s, now fast (in m/s}) is the boat approaching the dock when it is m from the dock? (Round your answer to two decima places _mfsNeed Help?Watch ItMaster [
boat is pulled into dock by pe attached to the bow of the boat and passing through pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at rate of m}s, now fast (in m/s}) is the boat approaching the dock when it is m from the dock? (Round your answer to two decima...
5 answers
Lot g(r) = r'_2x+7 . Evaluate oach d the ddlosial Hnlts (a) Jimg6*) =(b) lim g() =(c) lim g(r)
Lot g(r) = r'_2x+7 . Evaluate oach d the ddlosial Hnlts (a) Jimg6*) = (b) lim g() = (c) lim g(r)...
5 answers
13 lc +o taak Lc atrroxhakbn of#t_knj fo) (~*)w"2 aF 2=1 (Fra"s) Tu 7m xto 97w"xesJko SGqq )
13 lc +o taak Lc atrroxhakbn of#t_knj fo) (~*)w"2 aF 2=1 (Fra"s) Tu 7m xto 97w"xesJko SGqq )...

-- 0.059742--