A 50. g sample of dry iron(II) sulfate has which mass if converted to the pentahydrate?

Answer is: mass of iron(II) sulfate pentahydrate (FeSO₄ · 5H₂O) would be 79,62 g.
m(FeSO₄) = 50,0 g.
m(FeSO₄ · 5H₂O) = ?
m(FeSO₄) : M(FeSO₄) = m(FeSO₄ · 5H₂O) : M(FeSO₄ · 5H₂O).
50 g : 151,9 g/mol = m(FeSO₄ · 5H₂O) : 241,9 g/mol.
m(FeSO₄ · 5H₂O) = 50 g · 241,9 g/mol ÷ 151,9 g/mol.
m(FeSO₄ · 5H₂O) = 79,62 g.
M - molar mass.

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Аноним Аноним · Answer 2 · 2019-10-15T12:11:07+02:00

Answer: The mass of [tex]FeSO_4.5H_2O[/tex] formed will be 79.9 grams.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)

Given mass of iron(II) sulfate = 50. g

Molar mass of iron(II) sulfate = 151.9 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }FeSO_4=\frac{50.g}{151.9g/mol}=0.33mol[/tex]

The chemical equation for the formation of [tex]FeSO_4.5H_2O[/tex] follows:

[tex]FeSO_4+5H_2O\rightarrow FeSO_4.5H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of iron (II) sulfate produces 1 mole of [tex]FeSO_4.5H_2O[/tex]

So, 0.33 moles of iron (II) sulfate will produce = [tex]\frac{1}[1}\times 0.33=0.33mol[/tex] of [tex]FeSO_4.5H_2O[/tex]

Now, calculating the mass of [tex]FeSO_4.5H_2O[/tex] by using equation 1, we get:

Molar mass of [tex]FeSO_4.5H_2O[/tex] = 242 g/mol

Moles of [tex]FeSO_4.5H_2O[/tex] = 0.33 moles

Putting values in equation 1, we get:

[tex]0.33mol=\frac{\text{Mass of }FeSO_4.5H_5O}{242g/mol}\\\\\text{Mass of }FeSO_4.5H_5O=(0.33mol\times 242g/mol)=79.9g[/tex]

Hence, the mass of [tex]FeSO_4.5H_2O[/tex] formed will be 79.9 grams.