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

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

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:

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

Calculating number of moles of $FeSO_4.7H_2O$

Number of moles of $FeSO_4.7H_2O=2\times n_{Fe_2O_3}=2\times 2.59\times 10^{-3}=5.18\times 10^{-3}mol$

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

Moles of $FeSO_4.7H_2O=5.18\times 10^{-3}mol$

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

Putting values in equation 1, we get:

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

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

Answer 2

Answer:

I doo not know this

Explanation:

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