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A solution contains an unknown mass of dissolved silver ions. When potassium chloride is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 245 mg. What mass of silver was in the original solution? (Assume that all of the silver was precipitated out of solution by the reaction.)

Respuesta :

Answer:

Mass of silver in the original solution = 0.1845 g

Explanation:

The precipitate must be of silver chloride. So,

Given that:

Mass of silver chloride = 245 mg = 0.245 g ( 1 mg = 0.001 g )

Molar mass of silver chloride = 143.32 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{0.245\ g}{143.32\ g/mol}[/tex]

[tex]Moles= 0.00171\ mol[/tex]

From the reaction,  

[tex]KCl+Ag^+\rightarrow AgCl+K^+[/tex]

1 mole of AgCl is formed from 1 mole of silver

So,

0.00171 mole of AgCl is formed from 0.00171 mole of silver

Moles of silver = 0.00171 moles

Molar mass of silver = 107.8682 g/mol

Mass = Moles*Molar mass = 0.00171 moles*107.8682 g/mol = 0.1845 g

Mass of silver in the original solution = 0.1845 g

jayilych4real

The mass of silver in the original solution = 0.1845 g

Calculations and Parameters:

We assume that the precipitate must be of silver chloride.

So,

Given that:

  • Mass of silver chloride = 245 mg = 0.245 g ( 1 mg = 0.001 g )
  • Molar mass of silver chloride = 143.32 g/mol

The formula for the calculation of moles is shown below:

Moles= Mass taken/Molar mass

Putting the values in:

Moles= 0.00171mol.

Given the reaction,

KCl + Ag^+ ----->AgCl + K^+

1 mole of AgCl is formed from 1 mole of silver

0.00171 mole of AgCl is formed from 0.00171 mole of silver

  • Moles of silver = 0.00171 moles
  • Molar mass of silver = 107.8682 g/mol

Mass = Moles*Molar mass

= 0.00171 moles*107.8682 g/mol

= 0.1845 g

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