Suppose 0.701g of iron(ii) chloride is dissolved in 50.ml of a 55.0mm aqueous solution of silver nitrate.
calculate the final molarity of chloride anion in the solution. you can assume the volume of the solution doesn't change when the iron(ii) chloride is dissolved in it.
be sure your answer has the correct number of significant digits.

Answer : The molarity of chloride anion in the solution is 0.05 mole/L

Explanation : Given,

Mass of = 0.701 g

Volume of solution = 50 ml = 0.050 L

Molarity of AgCl = 55.0 mM = 0.055 M

Molar mass of = 126.751 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.00275 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 2 moles of react to give 2 moles of

So, 0.00275 moles of react to give 0.00275 moles of

Now we have to calculate the molarity of .

As we know that, 1 mole of AgCl in solution gives 1 mole of silver ion and 1 mole of chloride ion.

So, the molarity of chloride ion = Molarity of AgCl = 0.05 mole/L

Therefore, the molarity of chloride anion in the solution is 0.05 mole/L

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