First slide
Stoichiometry and Stoichiometric Calculations
Question

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

(Ag=107.8, N=14,O=16,Na=23,Cl=35.5)

Difficult
Solution

16.9% solution of AgNO3 means 16.9 g of AgNO3 in 100 mL of solution.

16.9 g of AgNO3 in 100 mL solution = 8.45 g of AgNO3 in 50 mL solution. 

Similarly, 5.8% of NaCl in 100 mL solution = 2.9 g of NaCl in 50 mL solution.

The reaction can be represented as :

                      AgNO3+NaClAgCl+NaNO3

Initially       8.45/170        2.9/58.5

                =  0.049 mol    = 0.049 mol

Finally            0                        0           0.049 mol    0.049 mol

  Mass of AgCl precipitated =0.049×143.3

                                                                         =7.027 g

 

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