A.) During an experiment, a student adds 2.90 g of CaO to 400.0 mL of 1.500 M HC

Question

A.) During an experiment, a student adds 2.90 g of CaO to 400.0 mL of 1.500 M HCl. The student observes a temperature increase of 6.00 °C. Assuming the solution's final volume is 400.0 mL, the density if 1.00 g/mL, and the heat capacity is 4.184 J/(g·°C), calculate the heat of the reaction, Hrxn.

B.) During an experiment, a student adds 1.05 g of calcium metal to 200.0 mL of 0.75 M HCl. The student observes a temperature increase of 17.0 °C for the solution. Assuming the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g·°C), calculate the heat of the reaction, Hrxn.

C.) Write the balanced chemical equation for the standard enthalpy of formation of magnesium oxide. Include the phase labels. The arrow has been provided for you. Do not include charges.

Explanation / Answer

A) given volume of solution = 400ml, density of solution = 1 g/ml, mass of solution= volume* density = 400*1= 400 gm

specific heat is takes to be same as that of water= 4.184 J.gm.deg.c, temperature change = 6 deg.c

, since there is an increase in temperature, the reaction is exothermic and enthalpy change is -ve

enthalpy change=- mass of solution*specific heat*change in temperature =- 400* 4.184*6 joules =-10042 joules

moles of CaO= mass/molar mass = 2.9/56 =0.052 and moles of HCl= molarity* volume in L= 1.5*400/1000 =1.5*0.4=0.6 moles

the reaction between CaO and HCl is CaO+2HCl ------->CaCl2+ H2O

1mole of Cao requires 2 mole of HCl and hence theoretical ratio =1:2

actual ratio of CaO:HCl = 0.052 : 0.6 =1:11.53

so HCl is excess reactant and all the Cao gets consumed.

Enthalpy change = -10042/0.052 J/mole of CaO=-193915 J/mole of CaO.

B) mass of the solution =200*1 =200 gm, enthalpy change= -200*4.184* 17=-14226 joules

moles of Ca= 1.05/40=0.02625, moles of HCl = molarity* volume in L=0.75*200/1000= 0.15 moles

Ca+ 2HCl ------->CaCl2+ H2

as per the reaction, theoretical ratio of Ca: HCl =1:2, actual ratio of Ca: HCl =0.02625:0.15 = 1:8

hence here also Cao is limiting and HCl is excess. Hence enthalpy change= -14226/0.01875=-790333.3 J/mole

2. Mg(s)+ 0.5O2(g) ------->MgO (s), is the reaction for calculating the standard heat of formation of MgO.

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