The problem does not require simultaneous equations.
I would use 143.32 for molar mass AgCl and mols AgCl = 1.913/143.32 = 0.1334
mols NaCl = mols AgCl
g NaCl = mols NaCl x molar mass NaCl = 0.01334 x 58.44 = 0.7800 g
% = (0.7800g/0.8870g)*100 =87.94%
A 0.8870g sample of a mixture of nacl and nan03 is dissolved in water, and the solution is then treated with an excess of AGN03 to yield 1.913g of agcl. Calculate the percent by mass of the nacl in the mixture.
1.913g / 143 = 0.0133
x/58 ( 0.8870g -x) y/85 = 0.0133
0.01529 - 0.01724x + 0.01176x = 0.0133
x= 0.00199/0.00548
= 0.363/0.8870g x 100
=41%
100g - 41g = 59g
The right answer 87.87%
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