Question

A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 2.955 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3⋅xH2O, which was subsequently ignited to produce 0.201 g Fe2O3. What was the mass of FeSO4⋅7H2O in the 2.955 g sample?

Answer 1

Step-by-step explanation:

According to the reaction, 2 moles of
`FeSO_(4).7H_(2)O` =
`Fe_(2)O_(3)`

Molar mass of
`Fe_(2)O_(3)` is 159.69 g/mol and mass is given as 0.201 g.

Therefore, calculate the number of moles of
`Fe_(2)O_(3)` as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(0.201 g)/(159.69 g/mol)`

=
`1.25 * 10^(-3)` mol

Hence, moles of
`FeSO_(4).7H_(2)O` = 2 × moles of
`Fe_(2)O_(3)`

Therefore, moles of
`FeSO_(4).7H_(2)O` =
`2 * 1.25 * 10^(-3)` mol

= 0.0025 mol

Now, calculate the mass of
`FeSO_(4).7H_(2)O` as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

mass = no. of moles × molar mass of
`FeSO_(4).7H_(2)O`

=
`0.0025 mol * 278.01 g/mol`

= 0.695 g

Thus, we can conclude that 0.695 g is the mass of
`FeSO_(4).7H_(2)O` in 2.955 g of the given sample.