Question

`500 mL` of `0.2 M NaCl` sol. Is added to `100 mL` of `0.5 M AgNO_(3)` solution resulting in the formation white precipitate of `AgCl`. How many moles and how many grams of `AgCl` are formed? Which is the limiting reagent?

Answer 1

`underset({:(500 xx 0.2),(= 100 mmol):})(NaCl) + underset({:(100 xx 0.5),(= 50 mmol):})(AgNO_(3)) rarr AgCl + NaNO_(3)`
50 mmol of `AgNO_(3)` will react with 50 mmol of `NaCl` and
50 mmol of `AgCl` will be fromed.
`AgCL = 50 m "moles" = 50 xx 10^(-3) = 0.05 mol`
Weight of `AgCl = 0.05 xx 143.5 = 7.175 g`
Hence, `AgNO_(3)` is the limiting reagent.