What volume of 0.250 mol/L hydrochloric acid is nuetralized by 1.50g of calcium hydroxide?
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To determine the volume of hydrochloric acid neutralized, we can use the stoichiometry of the balanced chemical equation between hydrochloric acid (HCl) and calcium hydroxide (Ca(OH)2).
The balanced chemical equation is as follows:
2HCl + Ca(OH)2 -> CaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Ca(OH)2. This means the mole ratio between HCl and Ca(OH)2 is 2:1.
First, we need to calculate the number of moles of Ca(OH)2. We can use the formula:
moles = mass / molar mass
The molar mass of Ca(OH)2 is:
40.08 g/mol (for Ca) + 2 * (1.01 g/mol (for H) + 16.00 g/mol (for O)) = 74.10 g/mol
Therefore, the moles of Ca(OH)2 can be calculated as:
moles = 1.50 g / 74.10 g/mol = 0.0202 mol
Since the mole ratio between HCl and Ca(OH)2 is 2:1, the number of moles of HCl is twice that of Ca(OH)2:
moles of HCl = 2 * 0.0202 mol = 0.0404 mol
Finally, let's calculate the volume of HCl using the given concentration of 0.250 mol/L (or 0.250 mol/L * 1000 mL/L = 250 mol/m^3):
volume = moles / concentration
volume = 0.0404 mol / 0.250 mol/L = 0.1616 L = 161.6 mL
Therefore, the volume of 0.250 mol/L hydrochloric acid neutralized by 1.50 g of calcium hydroxide is 161.6 mL.