We can see here that the maximum volume of pure water that could be produced is approximately 5.62 cm³.
To find the maximum volume of pure water produced in the decomposition of hydrated zinc sulfate (ZnSO₄·7H₂O), we'll use the given mass data.
First, calculate the mass of hydrated zinc sulfate used:
Mass of ZnSO₄·7H₂O = Mass of boiling tube + ZnSO₄·7H₂O - Mass of boiling tube
Mass of ZnSO₄·7H₂O = 54.46 g - 41.64 g
Mass of ZnSO₄·7H₂O = 12.82 g
Now, determine the moles of ZnSO₄·7H₂O:
Molar mass of ZnSO₄·7H₂O = 287 g/mol
Moles of ZnSO₄·7H₂O = Mass of ZnSO₄·7H₂O / Molar mass of ZnSO₄·7H₂O
Moles of ZnSO₄·7H₂O = 12.82 g / 287 g/mol ≈ 0.0446 mol
From the balanced chemical equation:
ZnSO₄·7H₂O(s) → ZnSO₄(s) + 7H₂O(l)
We see that 1 mole of ZnSO₄·7H₂O produces 7 moles of water.
Therefore, moles of water produced = 7 × Moles of ZnSO₄·7H₂O
Moles of water produced = 7 × 0.0446 mol ≈ 0.3122 mol
Now, convert moles of water produced to grams using the molar mass of water:
Molar mass of H₂O = 18 g/mol
Mass of water produced = Moles of water produced × Molar mass of H₂O
Mass of water produced = 0.3122 mol × 18 g/mol ≈ 5.62 g
Finally, as given in the question, 1 cm³ of pure water has a mass of 1.00 g.
Therefore, the volume of water produced = Mass of water produced in grams = 5.62 cm³
Hence, the maximum volume of pure water that could be produced is approximately 5.62 cm³.