What volume of 1.00M ca(oh)2 is needed to neutralize 250.0mL of 0.100M HCL?
2HCl + Ca(OH)2 ==> 2H2O + CaCl2
mols HCl = M x L = ?
mols Ca(ON)2 = 2 x mols HCl (look at the coefficients in the balanced equation).
Then M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2. You know M and mols, solve for L.Convert to mL if needed.
To determine the volume of 1.00M Ca(OH)2 solution needed to neutralize 250.0mL of 0.100M HCl, you can use the concept of stoichiometry.
First, we need to write the balanced chemical equation for the reaction between Ca(OH)2 and HCl:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
According to the balanced equation, one mole of Ca(OH)2 will react with two moles of HCl.
Next, we can calculate the moles of HCl in the given sample:
moles of HCl = concentration of HCl x volume of HCl
= 0.100M x 0.250L (convert mL to L)
= 0.0250 moles
Since the stoichiometry of the balanced equation is 1:2 (Ca(OH)2:HCl), we can say that 0.0250 moles of HCl will react with half as many moles of Ca(OH)2.
moles of Ca(OH)2 = 0.0250 moles / 2
= 0.0125 moles
Finally, using the molarity and moles of Ca(OH)2, we can calculate the volume of 1.00M Ca(OH)2 solution needed:
volume of Ca(OH)2 = moles of Ca(OH)2 / concentration of Ca(OH)2
= 0.0125 moles / 1.00M
= 0.0125 L
= 12.5 mL (since 1 L = 1000 mL)
Therefore, you would need 12.5 mL of 1.00M Ca(OH)2 solution to neutralize 250.0 mL of 0.100M HCl.
To find the volume of 1.00M Ca(OH)2 needed to neutralize 250.0mL of 0.100M HCl, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between Ca(OH)2 (calcium hydroxide) and HCl (hydrochloric acid) is:
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl, meaning there is a 1:2 mole ratio between Ca(OH)2 and HCl.
To begin, let's convert the given volume of HCl (in mL) to moles:
moles of HCl = (volume of HCl in liters) x (concentration of HCl in M)
= (250.0 mL / 1000 mL/L) x (0.100 mol/L)
= 0.0250 moles
Now, using the mole ratio from the balanced equation, we can determine the moles of Ca(OH)2 required:
moles of Ca(OH)2 = (moles of HCl) x (1 mole Ca(OH)2 / 2 moles HCl)
= 0.0250 moles x (1/2)
= 0.0125 moles
Finally, we can find the volume of 1.00M Ca(OH)2 needed using its concentration:
volume of Ca(OH)2 = (moles of Ca(OH)2) / (concentration of Ca(OH)2)
= 0.0125 moles / 1.00 mol/L
= 0.0125 L
= 12.5 mL
Therefore, 12.5 mL of 1.00M Ca(OH)2 is needed to neutralize 250.0 mL of 0.100M HCl.