During an experiment , a student adds 0.601g of calcium metal to 150.0 mL of 1.6

Question

During an experiment , a student adds 0.601g of calcium metal to 150.0 mL of 1.65 M HCl.                             

                                The student observes a temperature increase of 130.0 C for the solution.                             

                                Assuming the solution's final volume is 150.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g C), calculate the heat of the                                reaction.                             

                                Delta H = KJ/mol

( the answer is not 5439 nor 5461)

Explanation / Answer

moles of HCL = 1.65*150*10^-3 = 0.2475 moles

Moles of calcium = 0.601/40 = 0.015 moles

1 mole of Ca reacts with 2 moles of HCl

so,

Ca is limiting reagent,

heat of reaction = m*s*dT = 4.184*0.601*130/0.015 = 21793.06 J/mol = 21.8 kJ/mol

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