Hydrochloric acid !!!HELP!!! Thanks!?

Question

15.A 0.200 L solution of .950M hydrochloric acid is used to neutralize a 10.0 g sample of calcium hydroxide in a coffee-cup calorimeter. Calcium hydroxide solution is known as slaked lime and is used in agriculture to “sweeten” acidic soils. The temperature in the calorimieter was observed to rise from 20.0*Cto29.2*C during the course of the neutralization. Assume that the density of the hydrochloric acid solution is 1.00g/mL and that the specific heat capacity of the system is the same as that for water,4.18J/g-K.Assume that only the water is available to absorb or release heat and that the reaction is at constant pressure.
2HCI(aq)+Ca(OH)2(aq)------CaCI2(aq)+2H20(I)
a.What was the limiting reagent in the neutralization reaction? Provide support for your answer.
b.How many moles of excess reagent were left over?
c.If 5.1 g of calcium chloride were obtained after the neutralization was complete, what was the percent yield?
d.What was the enthalpy change for the system in the coffee-cup calorimeter?
e.What was the heat of reaction per mole of water produced for the neutrailization reaction?

Answer 1

Reaction: 2HCl + Ca(OH)2 --> CaCl2 + 2H2O

a)
Moles of HCl in 0.2L of 0.950M
0.95 moles in 1 L
? in 0.2L => 0.95*0.2/1 = 0.19 moles

Moles of Ca(OH)2 [Molar mass = 74 g.mol-1]
No of moles = mass / molar mass = 10 / 74 = 0.135 moles

Now from the stoichiometry
1 mol of Ca(OH)2 requires 2 mol of HCl
0.135 mol of Ca(OH)2 require ? ==> 0.135*2/1 = 0.270 mol of HCl
However, you only have 0.19 mol of HCl, therefore the reaction will stop when the HCl is used up, and thus the HCl is the limiting reactant.


b) Since the HCl is the limiting reactant, the Ca(OH)2 is the reagent in excess.

From stoichiometry:
2 mol of HCl react with 1 mol of Ca(OH)2
0.19 mol of HCl react with ? ==> 0.19*1/2 = 0.095 mol

Since you initially had 0.135 mol of Ca(OH)2 [ans (a)]
Then moles of excess reagent left over = 0.135 - 0.095 = 0.04 mol


c)
From stoichiometry:
2 mol of HCl produce 1 mol of CaCl2
0.19 mol of HCl produce ? ==> 0.19*1/2 = 0.095 mol of CaCl2

Moles of CaCl2 produced [Molar mass: 111 g mol-1]
Moles = mass / molar mass = 5.1 / 111 = 0.046 mol of CaCl2

So you should have produced 0.095 mol, but you only produced 0.046 mol.
A yield of 0.095 mol would be 100% yield
A yield of 0.046 mol is ? ==> 0.046*100/0.095 = 48.4%


d) Based on the assumptions in the question, your volume is 0.2L, i.e. 200mL
Since the density is 1 g/mL, then the mass of liquid is 200 g.

deltaQ = m.c.deltaT = 200*4.18*[29.2-20.0] = 7691.2 J


e) From stoichiometry, 2 mol of HCl produce 2 mol of H2O
So, 0.19 mol of HCl produce 0.19 mol of H2O

So 7691.2J were released in the production of 0.19 mol of H2O
? J would be released in the production of 1 mol?

7691.2*1/0.19 = 40480J = 40.48 kJ/mol

Answer 2

Hydrochloric acid is the answer of hydrogen chloride (HCl) in water. that's a tremendously corrosive, reliable mineral acid and has significant commercial makes use of. that's stumbled on needless to say in gastric acid. The compound hydrogen chloride has the formula HCl. At room temperature, that's a drab gasoline, which types white fumes of hydrochloric acid upon touch with atmospheric humidity. Hydrogen chloride gasoline and hydrochloric acid are significant in technologies and industry. The formula HCl is generally used to refer, rather misleadingly, to hydrochloric acid, an aqueous answer derived from hydrogen chloride.

Answer 3

Start by breaking it down! Write and balance your equation for the reaction to start with: HCl + Ca(OH)2 --> CaCl2 + H2O (you balance)

To determine limiting, you have to calculate number of moles of your reactants. You should know how to do this. Now that you know the number of moles, based on the balanced equation which one is going to run out first?

Using this info, calculate how much excess reactant you have by subtracting the number of moles used from toatl moles present.

Percent yield is found by changing grams of product to moles and then plug into %yield formula (theoretical - actual......) x 100

Enthalpy uses the temperature change data, along with your formula for q.

Use this calculation and your balanced equation to solve for heat of reaction.

Hope this helps!