A 0.500 L solution of 7.50 M hydrochloric acid is used to neutralize a 250.0 g sample of calcium hydroxide.?

Question

A 0.500 L solution of 7.50 M hydrochloric acid is used to neutralize a 250.0 g sample of calcium hydroxide. Calcium chloride, a chemical that is used to control highway dust, strengthen concrete mixes and add flavor to foods, is produced by this reaction. 2HCl + Ca(OH)2 --> CaCl2 +2H2O
a)How many moles of hydrochloric acid are present in this reaction?
b)How many moles of calcium hydroxide are present in this reaction?
c)Was there enough hydrochloric acid to neutralize all of the calcium hydroxide? Provide support for your answer.
d)How many moles of excess reagent were left over?
e)What is the theoretical yield of calcium chloride?
f)If 187 g of calcium chloride were obtained after the neutralization was complete, what was the percent yield?

acctually no I did not thats just one question out of about 20 on a worksheet.

Answer 1

Wow, this is a long one. The questions that you've been given allow you to go through it step by step.

a) C = n/V (concentration equals moles divide by volume)
n = C x V
n = (7.50 M)(0.500 L)
n = 3.75 mol HCl

b) n = m/M (moles equals mass divided by molar/molecular mass)
n = (250.0 g) / (40.078 + 2 x 15.9994 + 2 x 1.00794)
n = 3.374 mol Ca(OH)2

c) No, there is not enough HCl to neutralize all of the Ca(OH)2. According to the balanced equation, two moles of HCl are needed to neutralize one mole of Ca(OH)2. 6.748 moles of HCl would be needed to neutralize the given mass of Ca(OH)2.

d) The excess reagent is the Ca(OH)2.

3.75 mol HCl x (1 mol Ca(OH)2 / 2 mol HCl)
= 1.875 mol Ca(OH)2 used

3.374 mol - 1.875 mol = 1.499 mol Ca(OH)2 remaining.

e) 3.75 mol HCl x (1 mol CaCl2 / 2 mol HCl)
= 1.875 mol CaCl2 produced

1.875 mol CaCl2 x (40.078 + 2 X 35.4527)
=208.1 g CaCl2 produced (theoretically)

f)187 g / 208.1 g x 100%
=89.9 % yield

Answer 2

An antacid (calcium hydroxide) neutralizes stomach acid (hydrochloric acid). Ca(OH)2 + 2 HCl --> CaCl2 + 2 H2O