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You are given a 1.50 g mixture of sodium nitrate and sodium chloride. you dissolve this mixture into 100. ml of water and then add an excess of 0.500 m silver nitrate solution. the white solid produced from the reaction is filtered, dried and measured. the white solid has a mass of 0.641 g. calculate the percent, by mass, of sodium chloride in the original unknown mixture.

Respuesta :

When AgNO₃ solution is added to mixture of NaNO₃ and NaCl, it reacts with NaCl and gives precipitate of AgCl, which is white solid. The reaction takes place in 1:1 stoicheometric ratio as per the following balanced reaction:

NaCl + AgNO₃→ AgCl ↓ + NaNO₃. As per reaction 1 mole NaCl reacts with 1 mole AgNO₃ and gives  of solid precipitate of 1 mole AgCl.

0.641 g= [tex]\frac{0.641}{143.32}[/tex]= 4.47 X [tex]10^{-3}[/tex] mole of solid AgCl is produced. This indicates 4.47 X [tex]10^{-3}[/tex] mole NaCl =[tex]4.47 X 10^{-3}X58.5[/tex] g of Nacl reacts to give 0.641 g of AgCl. So, mass percentage of NaCl present in mixture=[tex]\frac{0.26}{1.5} X 100[/tex]= 17.33 %.

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