Question

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?

Answer 1

Mass of KNO3 in the original mix is 146.954 g

Step-by-step explanation:

mass of
`KNO_3` in original 254.5 mixture.

moles of
`BaSO_4 = (mass)/(Molecular\ Weight)`

moles of
`BaSO_4  = (68.3)/(233.38)`

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of
`BaCl_2 = mol* molecular weight`

`= 0.2926* 208.23`

= 60.92 g

the AgCl moles
`= (mass)/(Molecular\ Weight)`

`= (199.1)/(143.32)`

= 1.3891 mol of AgCl

note that, the Cl- derive from both,
`BACl_2 and NaCl`

so

mole of Cl- f NaCl
`= (1.3891) - (0.2926* 2) = 0.8039` mol of Cl-

mol of NaCl = 0.8039 moles

`mass = mol* Molecular\ Weight  = 0.8039 * 58 = 46.626\ g \ of \ NaCl`

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