To calculate the grams of anhydrous nickel 2 sulfate (NiSO4) produced from 8.753 heptahydrate (NiSO4·7H2O), we need to consider the molar mass of both compounds and the stoichiometry of the reaction.

The molar mass of NiSO4 is 154.75 g/mol, while the molar mass of NiSO4·7H2O is 280.86 g/mol.

The chemical equation for the conversion of NiSO4·7H2O to anhydrous NiSO4 is:

NiSO4·7H2O → NiSO4 + 7H2O

From this equation, we can see that one mole of NiSO4·7H2O produces one mole of NiSO4.

To calculate the moles of NiSO4·7H2O present in 8.753 g, we divide the mass by its molar mass:

Moles of NiSO4·7H2O = 8.753 g / 280.86 g/mol = 0.0312 mol

Since one mole of NiSO4·7H2O produces one mole of NiSO4, we know that 0.0312 mol of NiSO4 will be produced.

Now, we can calculate the mass of anhydrous NiSO4 produced by multiplying the moles by its molar mass:

Mass of NiSO4 = 0.0312 mol * 154.75 g/mol = 4.838 g

Therefore, 8.753 grams of nickel sulfate heptahydrate will produce 4.838 grams of anhydrous nickel 2 sulfate.