Question

Douglasite is a mineral with the formula 2KCl # FeCl2 # 2H2O. Calculate the mass percent of douglasite in a 455.0-mg sample if it took 37.20 mL of a 0.1000-M AgNO3 solution to precipitate all the Cl2 as AgCl.

Answer 1

Answer : The mass percent of Douglasite is 63.75 %

Explanation :

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

`\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3* \text{Volume of solution}=0.1mole/L* 0.03720L=3.72* 10^(-3)mole`

As we know that,

Moles of
`AgNO_3` = Moles of AgCl = Moles of
`Cl^-` =
`3.72* 10^(-3)mole`

The formula of Douglasite is,
`2KCl·FeCl_2·2H_2O`

The molar mass of Douglasite = 311.88 g/mole

Now we have to calculate the moles of
`Cl^-` atoms.

In the formula of Douglasite, there are 4
`Cl^-` atoms.

`\text{Moles of }Cl^-\text{ atom}=(3.72* 10^(-3)mole)/(4)=9.3* 10^(-4)mole`

Now we have to calculate the mass of Douglasite.

`\text{Mass of Douglasite}=\text{Moles of Douglasite}* \text{Molar mass of Douglasite}`

`\text{Mass of Douglasite}=(9.3* 10^(-4)mole)* (311.88g/mole)=0.29005g=290.05mg`

conversion used : (1 g = 1000 mg)

Now we have to calculate the mass percent of Douglasite.

`\text{Mass percent of Douglasite}=\frac{\text{Theoretical mass of Douglasite}}{\text{Given mass of Douglasite}}* 100=(290.05mg)/(455.0mg)* 100=63.75\%`

Therefore, the mass percent of Douglasite is 63.75 %