Question

A powder contains FeSO4⋅7H2OFeSO4⋅7H2O (molar mass=278.01 g/mol),(molar mass=278.01 g/mol), among other components. A 2.605 g2.605 g sample of the powder was dissolved in HNO3HNO3 and heated to convert all iron to Fe3+.Fe3+. The addition of NH3NH3 precipitated Fe2O3⋅xH2O,Fe2O3⋅xH2O, which was subsequently ignited to produce 0.413 g Fe2O3.0.413 g Fe2O3. What was the mass of FeSO4⋅7H2OFeSO4⋅7H2O in the 2.605 g2.605 g sample

Answer 1

I doo not know this

Step-by-step explanation:

dont mine me just need points

Answer 2

Answer: The mass of
`FeSO_4.7H_2O` present in the sample is 1.440 g

Step-by-step explanation:

The chemical equation follows:

`FeSO_4.7H_2O+HNO_3\rightarrow Fe^(+3)\\\\2Fe^(3+)+NH_3\rightarrow Fe_2O_3.xH_2O\\\\Fe_2O_3.xH_2O\rightarrow Fe_2O_3`

From the above equations, it is visible that number of moles of
`Fe_2O_3` is formed by half the number of moles of
`FeSO_4.7H_2O`

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of iron (III) oxide = 0.413 g

Molar mass of iron (III) oxide = 159.70 g/mol

Putting values in equation 1, we get:

`\text{Moles of iron (III) oxide}=(0.413g)/(159.7g/mol)=2.59* 10^(-3)mol`

Calculating number of moles of
`FeSO_4.7H_2O`

Number of moles of
`FeSO_4.7H_2O=2* n_(Fe_2O_3)=2* 2.59* 10^(-3)=5.18* 10^(-3)mol`

Calculating the mass of
`FeSO_4.7H_2O` using equation 1, we get:

Moles of
`FeSO_4.7H_2O=5.18* 10^(-3)mol`

Molar mass of
`FeSO_4.7H_2O` = 278.01 g/mol

Putting values in equation 1, we get:

`5.18* 10^(-3)mol=\frac{\text{Mass of }FeSO_4.7H_2O}{278.01g/mol}\\\\\text{Mass of }FeSO_4.7H_2O=(5.18* 10^(-3)mol* 278.01g/mol)=1.440g`

Hence, the mass of
`FeSO_4.7H_2O` present in the sample is 1.440 g