Question

Suppose 1.07g of iron(II) chloride is dissolved in 300.mL of a 14.0 M 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.

Answer 1

Answer : The molarity of chloride anion in the solution is 0.003318 mole/L

Explanation : Given,

Mass of
`FeCl_2` = 0.701 g

Volume of solution = 300 ml = 0.3 L

Molarity of AgCl = 14.0 M = 14.0 M

Molar mass of
`FeCl_2` = 126.751 g/mole

First we have to calculate the moles of
`FeCl_2`.

`\text{Moles of }FeCl_2=\frac{\text{Mass of }FeCl_2}{\text{Molar mass of }FeCl_2}=(0.701g)/(126.751g/mole)=0.00553moles`

Now we have to calculate the moles of
`AgNO_3`.

`\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3* \text{Volume of solution}=14.0mole/L* 0.3L=4.2mole`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`FeCl_2(aq)+2AgNO_3(aq)\rightarrow 2AgCl(s)+Fe(NO_3)_2(aq)`

From the balanced reaction we conclude that

As, 1 moles of
`FeCl_2` react with 2 mole of
`AgNO_3`

So, 0.00553 moles of
`FeCl_2` react with
`0.00553* 2=0.01106` moles of
`AgNO_3`

From this we conclude that,
`AgNO_3` is an excess reagent because the given moles are greater than the required moles and
`FeCl_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`AgCl`.

As, 1 moles of
`FeCl_2` react to give 2 moles of
`AgCl`

So, 0.00553 moles of
`FeCl_2` react to give
`0.00553* 2=0.01106` moles of
`AgCl`

Now we have to calculate the molarity of
`AgCl`.

`\text{Molarity of }AgCl=\frac{\text{Moles of }AgCl}{\text{Volume of solution}}`

`\text{Molarity of }AgCl=(0.01106mole)/(0.3L)=0.003318mole/L`

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 = 0.003318 mole/L

Therefore, the molarity of chloride anion in the solution is 0.003318 mole/L