1.475-g sample containing NH4Cl(FM53.491),K2CO3(FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate,Na+B(C6H5)4-, to precipitateK+and

NH4+ions completely:

(C6H5)4B-+K+(C6H5)4BK(s)FM358.33(C6H5)4B-+NH4+(C6H5)4BNH4(s)FM337.27

The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all theNH3.

NH4++OH-NH3(g)+H2O

It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent ofNH4ClandK2CO3in the original solid.

Short Answer

Expert verified

The weight percentage ofm(NH4Cl) is 14.6%

The weight percentage of m(N2CO3)is 14.5%

Step by step solution

01

Calculate x and y:

Let consider,

X as m(NH4Cl)and y as m(K2CO3)

25 ml of the sample gave 0.617 g of precipitate which contained both of the products

14xM(NH4Cl)×M((C6H5)4BNH4)+(2y)M(K2CO3)×M((C6H5)4BK)=0.617g14x53.492×337.27+2y138.21×358.33=0.617g

Calculate for y we get,

122yM(K2CO3)×M((C6H5)4BK))=0.554g122y138.21×358.33=0.554gy=0.2137g

Calculate for x we get,

14x53.492×337.27+2×0.2137g138.21×358.33=0.617gx=0.2157g

02

Calculate the weight percentages:

wt%(x)=xm(sample)wt%(x)=0.2157g1.475gwt%(x)=14.6%

The weight percentage ofm(NH4Cl)is 14.6%

role="math" localid="1663664775980" wt%(y)=ym(sample)wt%(y)=0.2137g1.475gwt%(y)=14.5%

The weight percentage of m(K2CO3)is 14.5%

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Most popular questions from this chapter

A solid mixture weighing 0.5485 gcontained only ferrous ammonium sulfate hexahydrate and ferrous chloride hexahydrate. The sample was dissolved in 1MH2SO4 , oxidized to Fe3+ with H2O2, and precipitated with cupferron. The ferric cupferron complex was ignited to produce 0.1678 gof ferric oxide, Fe2O3(FM 159.69). Calculate the weight percent of Clin the original sample.

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Statistics of coprecipitation. 17In Experiment 1,200.0mL of solution containing 10.0mg of SO42-(from Na2SO4) were treated with excess Bacl2 solution to precipitate BaSO4 containing some coprecipitated Cl-. To find out how much coprecipitated was present, the precipitate was dissolved in 35mL of 98wt%H2SO4 and boiled to liberate , which was removed by bubbling gas through the H2SO4. The HCI/N2 stream was passed into a reagent solution that reacted with to give a color that was measured. Ten replicate trials gave values of 7.8,9.8,7.8,7.8,7.8,7.8,13,7,12.7,13.7, and 12.7. Experiment 2 was identical to the first one, except that the mL solution also contained of from ). Ten replicate trials gave 7.8,10,8,8.8,7.8,6.9, 8.8, 15.7 , 12.7 , 13.7and 14.7μmolCl-.

(a) Find the mean, standard deviation, and 95% confidence interval for Cl-in each experiment.

(b) Is there a significant difference between the two experiments? What does your answer mean?

(c) If there were no coprecipitate, what mass of BaSO4(FM 233.39) would be expected?

(d) If the coprecipitate is (FM 208.23), what is the average mass of precipitate(BaSO4+BaCl2)in Experiment 1. By what percentage is the mass greater than the mass in part (c)?

Propagation of error. A mixture containing only silver nitrate and mercurous nitrate was dissolved in water and treated with excess sodium cobalticyanide, Na3[Co(CN)6] to precipitate both cobalticyanide salts:

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