Question

200 ml of 3 M AgNO3 and 300 ml of 2 M AgNO.
are added to a solution containing 200 ml of 10 M
NaCl. The amount of NaCl left unprecipitated is
(1) 46.8 g
(2) 70.2 g
(3) 33.4 g
(4) 63.4 g​

Answer 1

Answer:

46.8g

Explanation:

3*200+2*300=M*500

M=1200/500

M=22/5

AgNO3+NaCl ----> AgCl +NaNO3

no. of moles of AgNO3=12/5*500/1000 = 1.2mole

no. of moles of NaCl =10*200/1000 =2 mole

here the given amount of AgNO3 is totally consumed ,so it is the limiting

reagent.

amount of NaCl left=0.8* 51.5

=46.8g

Hope it helps !