Question

Be sure to show all of your work and when performing calculations, report answers to the correct number of significant figures. There should be four questions on this pre-lab assignment. The specific heat of water is 4.184 J/(g oC).

Question:1
Using Hess's Law, show how the three reactions in the BACKGROUND section of the lab can be combined to obtain the standard molar enthalpy of formation of MgO(s),
Mg(s) + ½ O2(g) → MgO(s).

You won't have actual numbers to use here, you are just showing how the equations for the reactions and the enthalpy changes for those reactions can be manipulated to give the desired reaction.






Question:2
The equation you will be using this week for your calorimetry calculations is: -n ΔHo = Ccal Δ T + (m)(c)(ΔT). See the lab for an explanation of terms.

Given: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ; ΔHo = -57.7 kJ

20.5 mL of 1.10 M HCl is mixed with 29.5 mL of 1.08 M NaOH. The temperature of all solutions was 22.06oC prior to mixing. The final temperature of the reaction mixture was 24.50oC. What is the heat capacity of the calorimeter (calorimeter constant)? You can assume that volumes are additive and the density of the solution is the same as the density of water, 1.00 g/mL. See the lab for the specific heat of this solution. Be careful with units.






Question:3
If 0.21 grams of MgO(s) are combined with 79.1 mL of 1.09 M HCl, what is the limiting reactant? Write the reaction as part of your answer and show your work.

Following the laboratory procedure, what is ΔHo for the reaction if your calorimeter constant is 16 J/oC and ΔT was 12.3oC? You can assume that the density of the solution is the same as the density of water, 1.00 g/mL. See the lab for the specific heat of this solution.






Question:4
If you add 1.13 kJ of heat to 298 g of water at 49.1oC, what temperature change would you observe in the water and what would be the final temperature of the water? The specific heat capacity of water is 4.184 J/(g oC).

Answer 1

1)

the three reactions are

a) Mg (s) + 2HCl (aq) -----> MgCl2 (aq) + H2 (g)     -----> dHa

b) MgO (s) + 2 HCl (aq) -----> MgCl2 (aq) + H20 (l) -----> dHb

c) H2 (g) + 0.5 02 (g) ----> H20 (l)   ------> dHc

the heat of formation of Mgo is obtained by

dHfo MgO = dHa - dHb + dHc

2)

now

total volume = 20.5 + 29.5

total volume = 50

given density = 1

so

mass = volume

so

mass of solution = 50 g

also

specific heat of solution = 4.18

now

moles of HCl = molarity x volume (L)

moles of HCl = 1.1 x 20.5 x 10-3 = 22.55 x 10-3

we get

57.7 x 1000 x 22.55 x 10-3 = Ccal x ( 24.50 - 22.06) + ( 5032 x 4.18 x ( 24.50 -22.06)

solving

we get

Ccal = 323.07

so

heat capacity of calorimeter is 323.07 J / C

3)

we know that

moles = molarity x volume (L)

so

moles of HCL = 1.09 x 79.1 x 10-3

moles of HCL = 0.086

now

moles = mass / molar mass

so

moles of MgO = 0.21 / 40 = 5.25 x 10-3

now

the reaction is

MgO + 2HCl ----> MgCl2 + H20

so

moles of HCl required = 2 x moles of MgO

= 2 x 5.25 x 10-3

= 0.0105

but

0.086 moles of HCl is present

so

HCl is in excess

and

MgO is the limiting reagent

dH = 16 x 12.3

dH = 196.8 J

4)

we know that

heat = mass x specific heat capacity x temp change

Q = m x s x dT

so

1.13 x 1000 = 298 x 4.184 x ( T - 322.1)

solving

we get

T = 323 K

T = 50 C

so

the final temperature is 50 C