Question

Question #2976c

Answer 1

0.573g of silver chloride will be obtained.

Explanation:

Equation: `AgNO_3` + HCl `rarr` AgCl + `HNO_3`

As HCl is being added in excess, `AgNO_3` will be the limiting reagent.

Amount of `AgNO_3` present = 0.68g `-:` (107.87+14.007+15.999x3)
= 0.004002966 mol

`AgNO_3` `-=` AgCl

Amount of AgCl obtained = 0.004002966 mol

Mass of AgCl obtained = 0.004002966 mol x (107.87 + 35.45) = 0.573705216 g = 0.573g (3 s.f.)