What is the mass of the precipitate formed when 50ml of 16.9% solution of AgNO3 is mixed with 50 ml of 5.8% solution of NaCl?

Question

It's a question from chemistry for neet of chapter 1, some basic concepts of chemistry of class 11? 

Answer 1

The balanced equation for this precipitation reaction is

AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq)

Approach is to find moles of AgNO3 and moles of NaCl present.  Then determine which reactant is limiting. And finally, find the moles and grams of the precipitate, AgCl that is formed.

Moles AgNO3:  16.9% = 16.9g/100ml x 50 ml = 8.45 g x 1 mole/170 g = 0.0497 moles AgNO3

Moles NaCl:  5.8% = 5.8g/100 ml x 50 ml = 2.9 g x 1 mole/58.4 g = 0.0497 moles NaCl

Since they are present in equal amounts and the mole ratio in balanced eq. is 1:1, neither is limiting.

Moles AgCl produced = 0.0497 moles

Mass of AgCl produced = 0.0497 moles x 143g/mole = 7.1 grams (2 sig. figs.)