Question

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

Answer 1

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

Step-by-step 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=(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}* 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* 242g/mol)=79.9g`

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

Answer 2

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.