Question

Solution of 7.8g common salt is dissolved in a solution of 30g AgNO3. How much silver chloride will be formed and what will be the mass of residue.

Answer 1

Hence residue will be 7.34g and AgCl formed 19.13g

Answer 2

Answer: 19.13 g of silver chloride is formed and 7.33 g of AgNO₃ is left as residue.

Explanation:

AgNO₃ + NaCl  →  AgCl ↓  + NaNO₃

The reaction between silver nitrate and sodium chloride (common salt) is a precipitation reaction.

AgCl is not soluble in water and precipitates out as a white solid.

Molar mass of AgNO₃ = 169.87 g

Molar mass of NaCl  =  58.44 g

Molar mass of AgCl  = 143.32 g

Mass of AgNO₃ required to react with 58.44 g NaCl = 169.87 g

Mass of AgNO₃ required to react with 1 g NaCl =  $\frac{169.87}{58.44}$ g

Mass of AgNO₃ required to react with 7.8 g NaCl =  $\frac{169.87}{58.44} * 7.8$  g

Mass of AgNO₃ required to react with 7.8 g NaCl = 22.67 g

∴ AgNO₃ is taken in excess and it will be residue and NaCl will act as limiting reagent.

Mass of residue = Taken mass of AgNO₃ - Used mass of AgNO₃

Mass of residue =  30 g  - 22.67 g

Mass of residue =  7.33 g

Mass of AgCl formed from 58.44 g NaCl = 143.32 g

Mass of AgCl formed from 1 g NaCl = $\frac{143.32}{58.44}$ g

Mass of AgCl formed from 7.8 g NaCl =  $\frac{143.32}{58.44} * 7.8$ g

Mass of AgCl formed from 7.8 g NaCl =  19.13 g

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