Standard Heat of Formation Name Date Section Partner (Unless otherwise noted, 2

Question

Standard Heat of Formation Name Date Section Partner (Unless otherwise noted, 2 pts. for each blank and box) Trial 1 Trial 2 Volume or30 M HCI 80-Om Mass of Mg Mass ofa» M Ha /11-24. 0:41619 (calculated) Total Mass 161671 (ealtculated)1472g (ealoieds Mole of HCI 3.613sk Meakculated) Mole of Mg 2 (calculated) (calculated) Limiting reagent (1 pt. each) 2/2 .(1 pt, each) itt-21- Initial Temperature C. 400°C- Final Temperature (1 pt. each) (3 pts. each) Average Hrxn l Show calculations for H rxn 1 for Trial 1 . Trial 2 Trial 1 (1 pt. each) Volume of20 M HCI 84.0ml Mass of MgO /7779 Mass of 3.0 M HCI Total Mass 84.Om 906409 0 pt cach) (1 pt. each) (calculated) (calculated) 1(calculated) (calculated) 63

Explanation / Answer

For trial 1

the moles of HCl will come from the statement, you say you have 80 ml of 2 M solution of HCl

Molarity = moles / volume

moles = Molarity * volume, volume is 80 ml or 0.08 Liters

moles = 0.08 * 2 = 0.16 moles

mass of Mg is = 0.4969 , molar mass of mg is 24.3g/gmol

moles of Mg = 0.4969 / 24.3 = 0.024 moles of Mg

Reaction between HCl and Mg is

2 HCl + Mg ==== H2 + MgCl2 as you can see you need 2 moles of HCl for every mole of Mg, you have 0.16 moles of HCl you theoretically need 0.08 moles of Mg but you have 0.024 less than the required, this means that you will consume all of the Mg and some HCl will remain unreacted so the Mg is your limiting reactant

Since no information about the density of the solution is provided then i will assume that the total mass you have provided is correct

To calculate the heat of reaction you have to apply

Enthalpy =- mass solution * heat capacity solution * (Tf - Ti)

for the heat capacity since the main component is water (solution) , you have approximately 5 grams of HCl and 0.5 grams of Mg then we can say that the heat capacity of water is the main contribution to the total heat capacity then we can say that the heat capacity of the solution is very close to the heat capacity of water 4.184 J / g K so

Enthalpy is = -119.69 * 4.184 * (40 - 21.3) = -9 364.64 Joules

For trial 2 everything is the very similar

moles of HCl = 0.085 * 2 = 0.17 moles of HCl

moles of Mg = 0.8 / 24 = 0.0333 moles of Mg

again the limiting reactant is the Mg

ENthalpy of reaction is = -127.45 * 4.184 * (44.2 - 21.2) = -12 264.76 Joules according to your data

Average enthalpy is - 10 814.7 Joules , the reaction is releasing heat then the values should be put negative

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