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Sati [7]
3 years ago
15

A student has 10 mL of a solution that might contain any or all of the following cations at 0.01 M concentrations: Mn2+, Ba2+, A

g+ , and Cu2+. Addition of 10 mL of 1 M HCl causes a precipitate to form. After the precipitate is filtered off, 1 M H2SO4 is added to the filtrate and another precipitate forms. What is the second precipitate
Chemistry
1 answer:
Ivanshal [37]3 years ago
7 0
<span>For the answer to the question above,
Addition of HCl causes a precipitate to form: 
Ag+ & Cl- -- AgCl a white ppt 


H2SO4 is added to the supernate another precipitate forms: 
Ba+2 & SO4-2 --> BaSO4 a white ppt 


a solution of NaOH is added to the supernatant liquid until it is strongly alkaline. No precipitate is formed: 
no Mn+2 & no Cu+2 which would have ppt as hydroxides</span>
I hope this helps you.
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Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

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