Final answer:
To neutralize the 45.0 mL of 1M HClO3 solution with a 0.20M Ca(OH)2 solution, you would need 0.0900 mol of Ca(OH)2.
Step-by-step explanation:
To determine the volume of the Ca(OH)2 solution that will neutralize the 45.0 mL of 1M HClO3 solution, we need to use the balanced chemical equation:
Ca(OH)2 + 2HClO3 -> Ca(ClO3)2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HClO3. So the number of moles of HClO3 in 45.0 mL is:
Moles of HClO3 = (45.0 mL)(1 L/1000 mL)(1 mol/L) = 0.0450 mol
Since the ratio of Ca(OH)2 to HClO3 is 1:2, we need twice as many moles of Ca(OH)2 to neutralize the HClO3. Therefore, the number of moles of Ca(OH)2 needed is 2 * 0.0450 mol = 0.0900 mol.