What is the mass of the precipitate formed when 50 mL of 16.9 % solution of is mixed with 50 mL of 5.8% NaCl solution?
16.9% solution of means 16.9 g of in 100 mL of solution.
16.9 g of in 100 mL solution = 8.45 g of 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 :
Initially 8.45/170 2.9/58.5
= 0.049 mol = 0.049 mol
Finally 0 0 0.049 mol 0.049 mol