Name Section Experiment 14 Advance Study Assignment: Heat Effects and Calorimetr

Question

Name Section Experiment 14 Advance Study Assignment: Heat Effects and Calorimetry 1. A metal sample weighing 124.10 g and at a temperature of 99.3°C was placed in 42.87 g of water in a calorimeter at 239°C. At equilibrium the temperature of the water and metal was 41.6 C. a. what was Ar for the water? (Arr b. What was AM for the metal? c. How much beat flowed into the water? joules d. Taking the specific beat of water to be 4IS calculate de specific heat of the metal, using joules sec e. What is the approximate molar mass of the metal? (se Eq. 4.) 2. When 5,12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5 C, the temperature of dhe solution went up to 49.8 C a. Is dissolution reaction exothemic? Why? b Calculate emo using Equation 1. joules c. Find AH for the Peaction as it occurred in the calorimeter (Eq. 5)

Explanation / Answer

1.

a. dT for metal = 99.3 - 41.6 = 57.7 oC

b. dT for H2O = 41.6 - 23.9 = 17.7 oC

c. Heat flow into the water = mCpdT

                                          = 42.87 x 4.18 x 17.7 = 3171.78 J

d. Specific heat Cp of metal = 3171.78/124.10 x 57.7

                                             = 0.443 J/g.oC

e. molar mass of metal = 55.845 g/mol (for Fe)

2. NaOH + H2O

a. The solution is exothermic : Yes, as heat is released. The final temperature is greater than the initial temperature

b. q(H2O) = 51.55 x 4.18 x (49.8 - 24.5) = 5451.62 J

c. dH = 5451.62 x 40/5.12 = 42.591 kJ/mol

Data sheet #5

Calculations

Pressure of column of water = 750.57 mmHg

For sample 1,

Pressure of dry Hydrogen gas = 750.57 - 19.8 = 730.77 mmHg = 0.961 atm

moles H2 = PV/RT = 0.961 x 0.03125/0.08205 x (273 + 22.9) = 0.00124 mol

mass of Mg ribbon = moles x molar mass = 0.00124 x 24.3 = 0.03 g

Similarly calculation for sample 2 can be done

1. If vapor pressure of water is not taken into account, the final mass of Mg would be higher as the pressure which is directly proportional to moles of H2 released would be higher.

2. If the strips were not clean and coated with MgO, not all Mg has reacted, lower volume of H2 would be released and thus final mass of Mg calculated would lower.

3. If Mg-Al slloy was used, we would end up with higher amount of H2 formed and thus higher moles of Mg present. Thus final mass of Mg calculated would also be greater than actual value.

Mg + 2HCl ---> MgCl2 + H2

2Al + 6HCl --> 2AlCl3 + 3H2

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.