Stoichiometry and Stoichiometric Calculations

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Question

What is the mass of the precipitate formed when 50 mL of 16.9 %  solution of ${\mathrm{AgNO}}_{3}$ is mixed with 50 mL of 5.8% NaCl solution?

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Solution

16.9% solution of ${\mathrm{AgNO}}_{3}$ means 16.9 g of $AgN{O}_{3}$ in 100 mL of solution.16.9 g of $AgN{O}_{3}$ in 100 mL solution = 8.45 g of ${\mathrm{AgNO}}_{3}$ 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 :                      ${\mathrm{AgNO}}_{3}+\mathrm{NaCl}⟶\mathrm{AgCl}+{\mathrm{NaNO}}_{3}$Initially       8.45/170        2.9/58.5                =  0.049 mol    = 0.049 molFinally            0                        0           0.049 mol    0.049 mol

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