What volumr 0.250 mol/L hydrochloric acid is neutralized by 1.50g of calcium hydroxide?
Ca(OH)2 + 2 HCl >> CaCl2+ 2 H20
so moles HCl = 2*moles calcium hydroxide
= 2*1.50/74
VolumeHCl=MolesHCl/concentration=
=2*1.50/(74*.250)
To find the volume of 0.250 mol/L hydrochloric acid neutralized by 1.50g of calcium hydroxide, we need to use the balanced chemical equation for the reaction between hydrochloric acid (HCl) and calcium hydroxide (Ca(OH)2), which is as follows:
2HCl + Ca(OH)2 -> CaCl2 + 2H2O
The balanced equation tells us that 2 moles of hydrochloric acid react with 1 mole of calcium hydroxide to form 1 mole of calcium chloride and 2 moles of water.
First, we need to convert the mass of calcium hydroxide to moles using its molar mass. The molar mass of calcium hydroxide (Ca(OH)2) is calculated as:
40.08 g/mol (Ca) + 2 * (1.01 g/mol (H) + 16.00 g/mol (O)) = 74.10 g/mol
Therefore, the number of moles of calcium hydroxide (Ca(OH)2) is:
1.50 g / 74.10 g/mol = 0.020 mol
Since the stoichiometry of the balanced equation tells us that 2 moles of hydrochloric acid react with 1 mole of calcium hydroxide, we can determine the number of moles of hydrochloric acid needed. In this case, it would be:
2 * 0.020 mol = 0.040 mol
Now, we can use the concentration of the hydrochloric acid solution (0.250 mol/L) to find the volume of hydrochloric acid solution needed to contain 0.040 moles of hydrochloric acid. This can be calculated as:
Volume (L) = Number of moles / Concentration (mol/L)
Volume = 0.040 mol / 0.250 mol/L = 0.16 L or 160 mL
Therefore, 0.250 mol/L hydrochloric acid is neutralized by 1.50g of calcium hydroxide in a volume of 160 mL.