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

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original  254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of$BaSO_4 = \frac{68.3}{233.38}$

                               = 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

                         $= 0.2926\times 208.23$

                         = 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

                          $= \frac{199.1}{143.32}$

                          = 1.3891 mol of AgCl

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

so

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

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