When 4.10 g of hydrated magnesium sulfate, MgSO₄·xH₂O, were heated strongly, 2.00 g of anhydrous magnesium sulfate were obtained - Leaving Cert Chemistry - Question c - 2011

Next, we calculate the number of moles of anhydrous magnesium sulfate (MgSO₄) using its molar mass:

1. Molar mass of MgSO₄ = 120 g/mol
2. Moles of MgSO₄ = ( \frac{\text{mass}}{\text{molar mass}} = \frac{2.00 , g}{120 , g/mol} = 0.01667 , mol )

Now, we can calculate the moles of water lost using the molar mass of water (H₂O):

1. Molar mass of H₂O = 18 g/mol
2. Moles of water = ( \frac{2.10 , g}{18 , g/mol} = 0.11667 , mol )

The relationship between the number of moles of anhydrous magnesium sulfate and the number of moles of water is represented by the equation:

$0.01667 \, mol \text{ (MgSO₄)} : x \times 0.11667 \, mol \text{ (H₂O)}$

From stoichiometry, we know that:

$\frac{0.11667}{0.01667} = x$

Calculating this gives:

$x = 7$