Question

1. For this reaction in parts A and B, calculate the heat of the reaction using the following:

qrxn= -(MC^T + Ccal^T)

Ccalorimeter = 11.6 J/1*C

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DATA FOR Part A.

Trial #	1	2	3
Mass of calorimeter	41.582 g	41.601 g	41.605 g
Mass of calorimeter+acid	68.857 g	66.832 g	70.596 g
Temperature(C) (acid)	22.0	21.6	21.8
Mass of weighboat	0.703 g	0.704 g	0.701 g
Mass of Mg+weighboat	0.768 g	0.752 g	0.754 g
Mass of Mg	0.065 g	0.048 g	0.053 g
Temperature (highest/(C))	30.8	30.0	29.8
Mass of calorimeter + contents	68.839 g	66.847 g	70.587 g

Data Analysis NOTE: Do not average any numbers until you have completed step 2 of the data analysis. 1. For the reactions in parts A and B, calculate the heat of the reaction using the following: qrxn=-(mcAT + Ccalorimeter AT) For Csp, use the specific heat of the mixture = 3.97 J/gºC. For the calorimeter constant, Ccalorimeter, use the value of Ccalorimeter from the previous lab (if done). 2. Convert the value of 7 to J/g of the limiting reactant and then to kJ/mol, In this case, this value in kJ/mol will represent AHexn. 3. You will need to include a third reaction for the formation of liquid water from hydrogen and oxygen gases. The heat of formation for this reaction is in the thermodynamic tables in your textbook. 4. Now use Hess's law: Write the three reactions (Mg + HCI, MgO + HCl, and the formation of water) so that they may be combined to form the reaction of interest, the formation of MgO from its elements. Remember, you can reverse reactions, double them or whatever is necessary. Start by writing the formation reaction for MgO and then make sure that the appropriate reactants and products are present. Use the appropriate AHrxn values to calculate the AH, for the formation of Mgo.

Answer 1

For Trial 1 AT = (30.8 - 22.0C = 8.2°C m = man of mg = 0.065g Esp = 3.975 grlach Ccalorimeter = 11.6 JOCH Inn = -(MCAT & Cear AT) = -(0.065g €3.97 Jgma) (8.2°c)] - [11.690(4 (8.2° C)] = -2.115 - 95.6945 = -97.8045 Por Trial 2 DT= (30.0 -21.6°C = 8.4°C ме и см, а м = 0.очее Cep = 3.97 Jaroc се еt. GT'1 Grun a -MCDT - Ceal DJ s 0.0989)(3.975913-1) (8.4-2)] - (11.69 °C ^) (8.4°C)] = -1.60075 - 97.445 = -99.0407J

Por Terial 3 Дте (24. - 21.9 t set е имам 1 0 = 0. 653 4 Ср 3.47 Тест ted = 1.6 TC- qrun = meat - Ceal AT = -(o.o53 С.41 с е - 11.6 5't-ce '')) = 1с 93 - 12.8T e-dч, Че3 рисом и Е47. gоч - 49. очо1 - 09. 1e3 але. 7157 5