Question

What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a
solution of 0,68 g of silver nitrate ?
(C = 35,5, Ag = 108, N = 14,0
16)​

Answer 1

Answer:

Explanation:

AgNO3​+HCl→AgCl+HNO3​

  1              1                      1            

 170                                  143.5

Mass of AgCl formed =143.5×0.34/170=0.287gm