I worked this earlier today but with different numbers.
3.00 mols NaCl
mols AgNO3 = M x L = 0.9L x 0.5M = 0.45
.........AgNO3 + NaCl ==> AgCl
I.......0.45.....................
add.............3.00............
C.......-0.45..-0.45......+0.45
E.........0.....2.55.......0.45
You can see from the equilibrium line that 0.45 mols AgCl will form, all of the AgNO3 will react, and you will have 2.55 mols NaCl left unreacted.
If 3.00 moles of NaCl were mixed with 0.9 L of 0.5 M of AgNO3, how many moles of solid would form? Include the formula for the solid.
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