Question

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

Answer 1

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.

Answer 2

Answer: The mass of $FeSO_4.5H_2O$ formed will be 79.9 grams.

Explanation:

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(II) sulfate = 50. g

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

Putting values in equation 1, we get:

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

The chemical equation for the formation of $FeSO_4.5H_2O$ follows:

$FeSO_4+5H_2O\rightarrow FeSO_4.5H_2O$

By Stoichiometry of the reaction:

1 mole of iron (II) sulfate produces 1 mole of $FeSO_4.5H_2O$

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

Now, calculating the mass of $FeSO_4.5H_2O$ by using equation 1, we get:

Molar mass of $FeSO_4.5H_2O$ = 242 g/mol

Moles of $FeSO_4.5H_2O$ = 0.33 moles

Putting values in equation 1, we get:

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

Hence, the mass of $FeSO_4.5H_2O$ formed will be 79.9 grams.