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Suppose that 25.0 mL of 0.440 M sodium chloride is added to 25.0 mL of 0.320 M silver nitrate. How many moles of silver chloride precipitate? What would be the concentrations of each of the ions in the reaction mixture after the reaction?​

Sagot :

The number of moles of silver chloride that will precipitate is 0.008 mole

From the question,

We are to determine the number of moles of silver chloride that will precipitate

First,

We will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

NaCl + AgNO₃ → AgCl + NaNO₃

This means

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride and 1 mole of sodium nitrate

Now, we will determine the number of moles of each reactant present

  • For sodium chloride (NaCl)

Concentration = 0.440 M

Volume = 25.0 mL = 0.025 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of NaCl present = 0.440 × 0.025

Number of moles of NaCl present = 0.011 mole

  • For silver nitrate (NaNO₃)

Concentration = 0.320 M

Volume = 25.0 mL = 0.025 L

∴ Number of moles of NaNO₃ present = 0.320 × 0.025

Number of moles of NaNO₃ present = 0.008 mole

From the balanced chemical equation,

1 mole of sodium chloride reacts with 1 mole of silver nitrate to produce 1 mole of silver chloride

Then,

0.008 mole of sodium chloride will react with the 0.008 mole of silver nitrate to produce 0.008 mole of silver chloride

∴ 0.008 mole of silver chloride will be produced

Hence, the number of moles of silver chloride that will precipitate is 0.008 mole

Learn more here: https://brainly.com/question/18434602

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