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mixter17
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The mass of the anhydrous compound that is formed when 51.8 g of magnesium sulfate heptahydrate is heated in an oven until all the water is driven off is 25,30 g.

The reaction for the removal of water from Magnesium Sulfate Heptahydrate (MgSO₄·7H₂O) is the following one

MgSO₄·7H₂O + Δ → MgSO₄ + 7H₂O

To calculate the mass of the anhydrous compound you'll need to use the molar masses and the molar coefficients of the reaction in the following conversion factor:

[tex]gMgSO_4=51,8gMgSO_47H_2O* \frac{1 mol MgSO_47H_2O}{246,4746 g MgSO_47H_2O}* \\ \frac{1molMgSO_4}{1mol MgSO_47H_2O} * \frac{120,366g MgSO_4}{1mol MgSO_4} =25,30 g MgSO_4[/tex]

If 51.8 g of  [tex]\rm \bold{ MgSO_4.7H_2O}[/tex] is heated until all the water is driven off. The mass of the anhydrous compound is 25.4g.

Given here,

The mass of [tex]\rm \bold{ MgSO_4.7H_2O}[/tex] is 51.8 g

Molar mass of [tex]\rm \bold{ MgSO_4.7H_2O}[/tex] is 246g/mol

Molar mass of [tex]\rm \bold{ MgSO_4}[/tex] is 120.4 g/mol

[tex]\rm\texttt{ Number of mol} = \frac{Given Mass }{Molar Mass} \\\\\rm \texttt{Number of mol} = \bold{ \frac{51}{246.5}}\\\\\rm \texttt{Number of mol }= \bold{ 0.21 mol}[/tex]

The mass of  [tex]\rm \bold{ MgSO_4.7H_2O}[/tex]

[tex]\rm \bold{\texttt{The mass of }MgSO_4= 0.21\times 120.4}\\ \\\rm \bold {\texttt{The mass of }MgSO_4=25.4 g}[/tex]

Hence, we can conclude that the mass of the anhydrous compound is 25.4g.

To know more about Molar mass, refer to the link:

https://brainly.com/question/22997914?referrer=searchResults

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