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.
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.