A 0.649-g sample containing only K2SO4(FM174.27)and (NH4)2SO4(FM132.14)was dissolved in water and treated withBa(NO3)2to precipitate allSO4-2asBaSO4(FM233.39). Find the weight percent ofK2SO4in the sample if 0.977 g of precipitate was formed.

Short Answer

Expert verified

The solution for the mass percent of potassium sulfate from the given sample is 61.1 %.

Step by step solution

01

Calculate the mass percent of potassium sulfate from the given sample

Given

0.977gofK2SO4

Consider x is mass of potassium sulfate and y is mass of(NH4)H2SO4

x+y=0.649g

xg174.27g/mol+yg132.14g/mol=0.977g233.39g/mol

x174.27is moles of K2SO4

y132.14is moles of(NH4)H2SO4

Substituting y=(0.649g-x)we get,

x=0.397g

02

Calculation of Mass percent.

The mass percent is calculated as follows,

mass%=0.397g0.649g×100%=61.1%

The mass percent of potassium sulfate is 61.1%

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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?

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