Determining the Final Solution Volume after Mixing Sulfuric Acid and Calcium Hydroxide

What happens when 45mL of 0.25mol/L sulfuric acid is mixed with 25mL of 0.58mol/L Calcium hydroxide?

Final answer: When 45mL of 0.25mol/L sulfuric acid is mixed with 25mL of 0.58mol/L calcium hydroxide, sulfuric acid is the limiting reactant, and will be completely consumed, leaving excess calcium hydroxide.

Explanation:

When 45mL of 0.25mol/L sulfuric acid (H2SO4) is mixed with 25mL of 0.58mol/L Calcium hydroxide (Ca(OH)2), a neutralization reaction occurs. The balanced equation for what happens when these solutions are mixed is: Ca(OH)2 + H2SO4 → CaSO4 + 2H2O To determine if the solutions will completely react, we first calculate the moles of each reactant: H2SO4: 0.045 L * 0.25 mol/L = 0.01125 mol Ca(OH)2: 0.025 L * 0.58 mol/L = 0.0145 mol Now looking at the stoichiometry of the balanced equation, we see that it takes 1 mole of H2SO4 to react with 1 mole of Ca(OH)2. Given that we have more moles of Ca(OH)2, H2SO4 is the limiting reactant, and it will be completely consumed. Thus, the reaction will produce 0.01125 mol of CaSO4 and leave some excess Ca(OH)2.
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