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 used to control highway dust, strengthen concrete mixes, and add flavor to foods – is produced by the reaction.

2 HCl + Ca(OH)2 → CaCl2 + 2 H2 O

a. How many moles of hydrochloric acid are present in the reaction? There are 0.500 L of 7.50 M hydrochloric acid.

b. How many moles of calcium hydroxide are present in the reaction? There are 250.0 g of calcium hydroxide.

c. Is there enough hydrochloric acid to neutralize all of the calcium hydroxide? Use your answer to part b to determine how many moles of HCl are needed to neutralize that much calcium hydroxide. Compare your answer to part a, and see if there is enough HCl to neutralize all of the calcium hydroxide.

d. How many moles of excess reagent (calcium hydroxide) are left over? Calculate how many moles of calcium hydroxide could be converted by the HCl available, then subtract from your answer to part b.

e. What is the theoretical yield of calcium chloride? Calculate the number of moles of calcium chloride produced from the HCl available (you already did this in part d), then convert that number to grams.

f. If 187 g of calcium chloride were obtained after the neutralization was complete, what was the percent yield? Use your answer to part e to see how many grams theoretically were supposed to be produced.

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 used to control highway dust, strengthen concrete mixes, and add flavor to foods – is produced by the reaction.

2 HCl + Ca(OH)2 → CaCl2 + 2 H2 O

a. How many moles of hydrochloric acid are present in the reaction? There are 0.500 L of 7.50 M hydrochloric acid.
moles = M x L

b. How many moles of calcium hydroxide are present in the reaction? There are 250.0 g of calcium hydroxide.
moles = grams/molar mass

c. Is there enough hydrochloric acid to neutralize all of the calcium hydroxide? Use your answer to part b to determine how many moles of HCl are needed to neutralize that much calcium hydroxide. Compare your answer to part a, and see if there is enough HCl to neutralize all of the calcium hydroxide.

d. How many moles of excess reagent (calcium hydroxide) are left over? Calculate how many moles of calcium hydroxide could be converted by the HCl available, then subtract from your answer to part b.

e. What is the theoretical yield of calcium chloride? Calculate the number of moles of calcium chloride produced from the HCl available (you already did this in part d), then convert that number to grams.

f. If 187 g of calcium chloride were obtained after the neutralization was complete, what was the percent yield? Use your answer to part e to see how many grams theoretically were supposed to be produced.
%yield = (187 g/theoretical yield)*100 = ??

a. To find the number of moles of hydrochloric acid present in the reaction, we can use the concentration of the hydrochloric acid solution and its volume. The given solution has a concentration of 7.50 M and a volume of 0.500 L.

We can use the formula:
moles = concentration × volume

Substituting the values, we get:
moles = 7.50 M × 0.500 L = 3.75 moles of HCl

Therefore, there are 3.75 moles of hydrochloric acid present in the reaction.

b. To find the number of moles of calcium hydroxide present in the reaction, we can use the molar mass of calcium hydroxide (Ca(OH)2) and the given mass of 250.0 g.

The molar mass of calcium hydroxide is:
40.08 g/mol (for calcium) + 2 × 16.00 g/mol (for oxygen) + 2 × 1.01 g/mol (for hydrogen) = 74.10 g/mol

We can use the formula:
moles = mass / molar mass

Substituting the values, we get:
moles = 250.0 g / 74.10 g/mol = 3.376 moles of Ca(OH)2

Therefore, there are 3.376 moles of calcium hydroxide present in the reaction.

c. To determine if there is enough hydrochloric acid to neutralize all of the calcium hydroxide, we compare the number of moles of HCl (from part a) to the number of moles of Ca(OH)2 (from part b).

From part a, we have 3.75 moles of HCl.
From part b, we have 3.376 moles of Ca(OH)2.

Since the stoichiometric ratio between HCl and Ca(OH)2 is 2:1 (from the balanced equation), we know that for every 2 moles of HCl, 1 mole of Ca(OH)2 is neutralized.

Therefore, the number of moles of HCl needed to neutralize 3.376 moles of Ca(OH)2 is:
(3.376 moles Ca(OH)2) × (2 moles HCl / 1 mole Ca(OH)2) = 6.752 moles of HCl

Since we only have 3.75 moles of HCl, there is not enough HCl to completely neutralize all of the calcium hydroxide.

d. To find the number of moles of excess reagent (calcium hydroxide) left over, we subtract the number of moles of Ca(OH)2 used for neutralization (6.752 moles from part c) from the initial number of moles of Ca(OH)2 (3.376 moles from part b).

Excess moles of Ca(OH)2 = 3.376 moles - 6.752 moles = -3.376 moles

Since we cannot have negative moles, it means that all of the calcium hydroxide is consumed, and there is no excess reagent left over.

e. The number of moles of calcium chloride produced from the available HCl is the same as the number of moles of HCl used for neutralization.

So, the moles of calcium chloride (CaCl2) produced is 6.752 moles (from part c).

To calculate the theoretical yield of calcium chloride in grams, we can use its molar mass, which is:
40.08 g/mol (for calcium) + 2 × 35.45 g/mol (for chlorine) = 111.98 g/mol

Theoretical yield (in grams) = moles of CaCl2 × molar mass of CaCl2
= 6.752 moles × 111.98 g/mol = 756.62 g

Therefore, the theoretical yield of calcium chloride is 756.62 grams.

f. The percent yield is the actual yield divided by the theoretical yield, multiplied by 100%.

Given that the actual yield is 187 g and the theoretical yield is 756.62 g (from part e), we can calculate the percent yield as follows:

Percent yield = (actual yield / theoretical yield) × 100%
= (187 g / 756.62 g) × 100%
= 24.7%

Therefore, the percent yield of calcium chloride is 24.7%.