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ID: 526827 • Letter: H

Question

http LOG IN FROM ANYWHERE User Name: FirstName LastName@s Password Same as campus compu e Home Insert Page Layout Formulas Data Review View Tell me what you want to do ean contain viruses Unless you need to edit safer to stay in Protected view. Enable Editi careful Hues famneinternet PROTECTED VIEW Be Print and tape this on to the lab notebook under Data section please. MUST use the CLASS AVERAGE for your analysis. Group Mgo HCI 1 0077 25.0 22.3 33.7 11.4 0.249 25.0 22.4 28.2 5.8 2 0.064 25.0 21.5 32.0 10.5 0.249 25.0 23.5 31.0 7.5 4 0.153 25.0 22.8 46.5 23.7 0.252 25.0 23.1 29.9 6.8 5 0.155 25.0 23.0 46.5 23.5 0.245 23.0 23.0 26.0 3.0 az 6 0.062 25.0 22.5 31.1 8.6 0.250 25.0 21.8 28.0 6.2 7 0.075 25.0 23.1 34.1 11.0 0.254 25.0 22.9 30.0 7.1 8 0.077 25.0 21.5 33.0 11.5 0.257 25.0 22.5 30.0 7.5 15 average 0.095 25.0 224 36 l 14.3 0.251 24.7 227 290 6.3 O Im Cortana. Ask me anything.

Explanation / Answer

A) Mg + 2HCl ---> MgCl2 + H2

A1. moles Mg

First run, moles Mg = 0.077/24.305 = 0.0032 mol

Second run, moles Mg = 0.064/24.305 = 0.0026 mol

A2. Limiting reactant = Mg

Mg is gaining heat, HCl is loosing heat

A3. enthalpy of reaction,

First run, dHrxn = 25.077 x 4.18 x 11.4/0.0032 = 373.43 kJ/mol

Second run, dHrxn = 25.064 x 4.18 x 10.5/0.0026 = 423.10 kJ/mol

similarly others can be calculated

B) MgO + 2HCl ---> MgCl2 + H2O

A1. moles Mg

First run, moles Mg = 0.249/24.305 = 0.0102 mol

Second run, moles Mg = 0.249/24.305 = 0.0102 mol

A2. Limiting reactant = MgO

MgO is gaining heat. HCl is loosing heat

A3. enthalpy of reaction,

First run, dHrxn = 25.249 x 4.18 x 5.8 = 60.01 kJ/mol

Second run, dHrxn = 25.249 x 4.18 x 7.5 = 77.60 kJ/mol

similarly others can be calculated

A.7. Hess's Law

Enthalpy of combustion of Mg

dHcomb from first run = 373.43 - 60.01 = 313.42 kJ/mol

dHcomb from second run = 423.10 - 77.60 = 345.5 kJ/mol