A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved in water and allowed to react with excess H2SO4, 68.3 g a white precipitate is collected. When the remaining 138.0 g of the mixture is dissolved in water and allowed to react with excess AgNO3, 199.1 g of a second precipitate is collected. What is the mass of KNO3 in the original 254.5 g mixture?

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rejkjavik rejkjavik · Answer 1 · 2019-10-05T07:38:09+02:00

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of [tex]KNO_3[/tex] in original 254.5 mixture.

moles of [tex]BaSO_4 = \frac{mass}{Molecular\ Weight}[/tex]

moles of[tex] BaSO_4 = \frac{68.3}{233.38}[/tex]

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of [tex]BaCl_2 = mol\times molecular weight[/tex]

[tex] = 0.2926\times 208.23[/tex]

= 60.92 g

the AgCl moles [tex]= \frac{mass}{Molecular\ Weight}[/tex]

[tex]= \frac{199.1}{143.32}[/tex]

= 1.3891 mol of AgCl

note that, the Cl- derive from both, [tex]BACl_2 and NaCl[/tex]

so

mole of Cl- f NaCl [tex]= (1.3891) - (0.2926\times 2) = 0.8039[/tex] mol of Cl-

mol of NaCl = 0.8039 moles

[tex]mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl[/tex]

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g