Question

what is the mass of the precipitate formed when 50 ml of 16.9% solution of AgNo^3 is mixed with 59 ml of 5.8%NaCL solution? ?

Answer 1

moles of AgNO3 = > 50 * 16.9/(100 * 170) = > 0.05 (approx.)
moles of NaCl = > 5.8 * 5.8/(100 * 58.5) => 0.05
The reaction goes like this:
AgNO3 + NaCl ---------> AgCl + NaNO3
0.05 0.05 0 0
0 0 0.05

mass of agcl = 0.05 * 143.5 => 7.16 g.
Hence this would be the final mass of the precipitate.