Question

		Used Mg for these Trials	Initial Temperature	Final Temperature
Trials	HCl (mL)	Mass (g)	TioC	TfoC
Trial #1	7.5 mL	0.08 g	24 oC	62 oC
Trial #2	7.5 mL	0.06 g	23.9 oC	56.9 oC

		Used MgO for these Trials
Trials	HCl (mL)	Mass (g)	TioC	Tf oC
Trial #1	7.5 mL	0.13 g	23.8 oC	30.1 oC
Trial #2	7.5 mL	0.16 g	22.7 oC	36.7 oC

Calculate the heat absorbed by the HCl for each trial. Assume that the specific heat capacity of the dilute HCl is the same as water(4.184 J/g∙°C)

Determine the limiting reagent and calculate the moles of the limiting reagent for each trial.

Calculate ∆Hrxn for each trial.

Calculate ∆Hf° for MgO then calculate % error using the actual value of ∆Hf° for MgO.

Answer 1

heat absorbed by HCL= mass of HCl* specific heat* temperature difference=

The reaction is Mg+2HCl--------->MgCl2+ H2

Trail-1 : mass of HCl= Volume*density = 7.5 ml*1g/ml= 7.5gm, specific heat= 4.184 J/gm.deg.c and temperature difference= 62-24= 38, heat absorbed = 7.5*4.184*38 joules =1192 joules

moles of Mg used = mass/molar mass= 0.08/24=0.00333, heat absorbed by Mg/mole= 1192/0.00333 = 357600 J/mole

Trial-2: heat absorbed= 7.5* 1*4.184*(56.9-23.9) joules/ 0.06/24 J/mole =414216 J/mole= 414.216 KJ/mole

since there is temperature rise, deltaH= -357.600 Kj/mole for trial-1 and 414.216 Kj/mole for trial-2

2. MgO+2HCl------->MgCl2+ H2P

molar mass of MgO= 40

Heat absorbed= 7.5* 1*(30.1-23.8)*4.184/(0.13/40)=60829 j/mole= 60.829 Kj/mole

since temperature rise is there, deltaH=-60.829 Kj/mole

Trial-2 = 7.5*1*4.184*(36.7-22.7)/ (0.16/40)=109830 J/mole= 109.830 KJ/mole

deltaH=-109.830 Kj/mole

for trial -1 % error= 100*{(-130.65+60.829)/-130.65}=53.44%

actual vallue is =-130.65 Kj/mole, % error=100*{ (-130.65+109.830)/-130.65}=15.9%

% error=