Question

If 36.0 g of mgso4⋅7h2o is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Answer 1

Answer is: mass of anhydrous magnesium sulfate 17,50 grams will remain.
m(MgSO₄·7H₂O) = 36,0 g.
m(MgSO₄) = ?
Make proportion: m(MgSO₄) : m(MgSO₄·7H₂O) = M(MgSO₄) : M(MgSO₄·7H₂O).
m(MgSO₄) : 36,0 g = 120,36 g/mol : 247,47 g/mol.
m(MgSO₄) = 17,50 g.

Answer 2

Answer: The mass of anhydrous magnesium sulfate that will remain is 17.6 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
`MgSO_4.7H_2O` = 36.0 g

Molar mass of
`MgSO_4.7H_2O` = 246.47 g/mol

Putting values in equation 1, we get:

`\text{Moles of }MgSO_4.7H_2O=(36.0g)/(246.47g/mol)=0.146mol`

The chemical equation for the heating of
`MgSO_4.7H_2O` follows:

`MgSO_4.7H_2O\rightarrow MgSO_4+7H_2O`

By Stoichiometry of the reaction:

1 mole of
`MgSO_4.7H_2O` produces 1 mole of anhydrous magnesium sulfate.

So, 0.146 moles of
`MgSO_4.7H_2O` will produce =
`(1)/(1)* 0.146=0.146mol` of anhydrous magnesium sulfate

Now, calculating the mass of anhydrous magnesium sulfate by using equation 1, we get:

Molar mass of anhydrous magnesium sulfate = 120.37 g/mol

Moles of anhydrous magnesium sulfate = 0.146 moles

Putting values in equation 1, we get:

`0.146mol=\frac{\text{Mass of anhydrous magnesium sulfate}}{120.37g/mol}\\\\\text{Mass of anhydrous magnesium sulfate}=(0.146mol* 120.37g/mol)=17.6g`

Hence, the mass of anhydrous magnesium sulfate that will remain is 17.6 grams.