Ag = 108 g/mol
Cl = 35.5 g/mol
AgCl = 108 + 35.5 = 143.5 g/mol
1.933/143.5 = .0135 mols of AgCl so used .0135 mols of Cl
.0135 * 35.5 = .478 grams of Cl used
Na = 23 g/mol so NaCl = 58.5 g/mol
K = 39 g / mol so KCl = 74.5 g/mol
say used x grams NaCl and (.9 - x) grams of KCl
x grams NaCl --> x/58.5 mols NaCl
(.9-x)grams KCl --->(.9-x)/74.5 mols KCl
BUT we know that .0135 mols Cl were used
SO x/23 + (.9-x)/74.5 = .0135
solve for x, the grams of NaCl, .9-x the grams of KCl etc
100 (x/.9) = percent Na
A 0.9000g sample of a mixture NaCl and KCl is dissolved in water and the solution is then treated with an excess of AgNO3 to yield 1.933g of AgCl. Calculate the percent by mass of each compound in the mixture
1 answer
1 answer