If 62.7 ml of .208M HCL is needed to needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?
Determine your moles of HCl. Then use your equation to convert moles of HCl to moles of Ca(OH)2 then to grams of the calcium hydroxide.
The ratio would be 2Mol HCL to 1mol of Ca(OH)2
So you get 0.01304 mol HCL.
YOU GET 0.483GRAMS OF CALCIUM HYDROXIDE.
To find the number of grams of Ca(OH)2 in the solution, we need to use the concept of stoichiometry and the balanced chemical equation of the reaction between HCl and Ca(OH)2.
The balanced chemical equation for the neutralization reaction is:
2HCl + Ca(OH)2 → CaCl2 + 2H2O
From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of Ca(OH)2.
First, let's calculate the number of moles of HCl used. We can use the formula:
moles = concentration (M) × volume (L)
Given that the volume of HCl is 62.7 ml (0.0627 L) and the concentration is 0.208 M, we can calculate:
moles of HCl = 0.208 M × 0.0627 L = 0.01303 moles
Since the stoichiometry ratio between HCl and Ca(OH)2 is 2:1, the number of moles of Ca(OH)2 is half the number of moles of HCl:
moles of Ca(OH)2 = 0.01303 moles / 2 = 0.006515 moles
Now, let's calculate the number of grams of Ca(OH)2 using the molar mass of Ca(OH)2. The molar mass of Ca(OH)2 is:
molar mass of Ca(OH)2 = (1 × atomic mass of Ca) + (2 × atomic mass of O) + (2 × atomic mass of H)
= (1 × 40.08 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol)
= 74.10 g/mol
Finally, we can calculate the mass of Ca(OH)2:
mass = moles × molar mass
= 0.006515 moles × 74.10 g/mol
≈ 0.482 grams
Therefore, approximately 0.482 grams of Ca(OH)2 must be present in the solution.