Question #2976c

1 Answer
Aug 19, 2017

Answer:

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.)