Question

Suppose 0.701g of iron(II) chloride is dissolved in 50.mL of a 55.0mM aqueous solution of silver nitrate.

Answer 1

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

Explanation : Given,

Mass of $FeCl_2$ = 0.701 g

Volume of solution = 50 ml = 0.050 L

Molarity of AgCl = 55.0 mM = 0.055 M

Molar mass of $FeCl_2$ = 126.751 g/mole

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

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

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

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=0.055mole/L\times 0.050L=0.00275mole$

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, 2 moles of $AgNO_3$ react with 1 mole of $FeCl_2$

So, 0.00275 moles of $AgNO_3$ react with $\frac{0.00275}{2}=0.001375$ moles of $FeCl_2$

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

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

As, 2 moles of $AgNO_3$ react to give 2 moles of $AgCl$

So, 0.00275 moles of $AgNO_3$ react to give 0.00275 moles of $AgCl$

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

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

$\text{Molarity of }AgCl=\frac{0.00275mole}{0.055L}=0.05mole/L$

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 = 0.05 mole/L

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