What is the minimum mass of anhydrous CuSO4 needed to remove 12.4g H2O from a tankful of gasoline?

To calculate the minimum mass of anhydrous CuSO4 required to remove water from gasoline, we need to determine the stoichiometric relationship between CuSO4 and H2O in the dehydration reaction.

The chemical formula for anhydrous copper(II) sulfate is CuSO4, which means it does not contain any water molecules within its structure. In contrast, the hydrated form CuSO4·nH2O contains a specific number of water molecules (n) per formula unit.

To remove water from gasoline, we can use the dehydration reaction:

CuSO4·nH2O → CuSO4 + nH2O

By balancing this equation, we find that for every mole of CuSO4·nH2O, one mole of water is released.

Now, let's determine the molar mass of CuSO4. Copper (Cu) has a molar mass of approximately 63.55 g/mol, sulfur (S) has a molar mass of approximately 32.07 g/mol, and oxygen (O) has a molar mass of approximately 16.00 g/mol. Combining these masses together, the molar mass of CuSO4 is:

63.55 g/mol (Cu) + 32.07 g/mol (S) + (4 × 16.00 g/mol) (O) = 159.61 g/mol

Next, calculate the molar mass of water (H2O):

2 × 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol

Now, we can calculate the number of moles of water in 12.4 g of H2O by dividing the given mass by the molar mass:

12.4 g H2O ÷ 18.02 g/mol ≈ 0.688 mol H2O

Since the reaction is 1:1, for every mole of water, we need one mole of CuSO4. Therefore, we require approximately 0.688 moles of CuSO4.

Finally, to calculate the minimum mass of anhydrous CuSO4 needed, we multiply the number of moles by the molar mass:

0.688 mol CuSO4 × 159.61 g/mol ≈ 110 g

Therefore, the minimum mass of anhydrous CuSO4 needed to remove 12.4 g of water from the gasoline tank is approximately 110 grams.