posted by keitanako
A 10.00 gram solid sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in 100.0 mL of water is combined with another 200.0 mL solution that contains excess aqueous lead(II) nitrate. Upon mixing the two solutions, a solid precipitate forms (it is a mixture of two ionic solids.) It is filtered, dried, and found to have a dry mass of 21.75 grams. Determine the mass percent of sodium chloride in the original solid sample mixture of sodium chloride and potassium sulfate.
Two equations in two unknowns; solve simultaneously.
Let X = mass NaCl
and Y = mass K2SO4
eqn 1 is X + Y = 10.0
eqn 2 is mass PbCl2 (from NaCl) + mass PbSO4(from K2SO4) = 21.75g.
eqn 2 in terms of X and Y is this
(X*MMPbCl2/2*MMNaCl) + (Y*MMPbSO4/MMK2SO4) = 21.75
Solve those two equations for X and Y, then
%NaCl = (Xgrams/10)*100 = ?
Hi, Bob. But how to get (MMPbCl2/2*MMNaCl) and (MMPbSO4/MMK2SO4). Because it is confusing to me. Can you show the step more clearly. Thank you?
I just can do step 1, which is X=10-Y.
But step 2, it is bit confused. I can't get the meaning.
Thank you, I know how to do it.