(CHEMISTRY: Stoichiometry) Did i get the correct answers

Ca(OH)2(s) + 2 HCl(aq) ---> CaCl2(aq) + 2 H2O(l)

Using this reaction:
a) How many liters of 2.00M Hcl would be required to react completely with 2.54 grams of calcium hydroxide? ((0.0343L HCl))

b) How many grams of calcium hydroxide are neede to react with 69.50mL of 0.350M hydrochloric acid? ((0.901g Ca(OH)2))

c) If 350.0mL of 0.250M calcium chloride were produced, what mass of calcium hydroxide was reqUIRED? (assume an excess of HcL) ((6.48g Ca(OH)2 ))

I put my answers in ((brackets)) . THANKS! :)

a.) 2.54-g Ca(OH)2/74.08-g/mol X (2 mol HCl / 1 mol Ca(OH)2 = 0.0686 mol HCl --->
M = mol/L ---> 2.0M = .0686-mol/L ---> V = 0.0343-L Good job!

b.) 69.50-mL of 0.350M HCl = .350mol/L x 69.50-mL x (1-L / 1000-mL) = 0.0243 mol HCl--->
0.0243 mol HCl x 1 mol Ca(OH)2 / 2 mol HCl x 74.08-g Ca(OH)2 / 1 mol Ca(OH)2 = 0.901-g You got that one too!

c.) 350-mL = 0.350-L---> 0.250mol/L x 0.350-L = 0.0875 mol CaCl2--->
0.0875-mol CaCl2 x (1mol Ca(OH)2 / 1 mol CaCl2) x ( 74.08-g /1 mol Ca(OH)2 = 6.48-g Ca(OH)2
Excellent job! You got them all correct!

a) moles HCl required = (2.54 grams Ca(OH)2) * (1 mol Ca(OH)2 / 74.093 grams Ca(OH)2) * (2 mol HCl / 1 mol Ca(OH)2) = 0.0686 moles HCl required

molarity = moles solute / liters solution

2.00 M = 0.0686 moles HCl / liters solution

liters solution = 0.0343 Liters is the answer.

You are correct.

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(b) molarity = moles solute / liters solution

0.350 M HCl = moles HCl / 0.06950 Liters

moles HCl = 0.024325 moles HCl

grams Ca(OH)2 required = (0.024325 moles HCl) * (1 mol Ca(OH)2 / 2 mol HCl) * (74.093 grams Ca(OH)2 / 1 mol Ca(OH)2) = 0.901 grams Ca(OH)2 is the answer.

You are correct.

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(c) molarity = moles solute / liters solution

0.250 M CaCl2 = moles CaCl2 / 0.350 Liters

moles CaCl2 = 0.0875 moles CaCl2

mass Ca(OH)2 required = (0.0875 moles CaCl2) * (1 mol Ca(OH)2 / 1 mol CaCl2) * (74.093 grams Ca(OH)2 / 1 mol Ca(OH)2) = 6.483 grams Ca(OH)2 is the answer.

You are correct.