Chemistry Problem

Question

During an experiment, a student adds 0.601 g of calcium metal to 150.0 mL of 1.65 M HCl. The student observes a temperature increase of 13.0 °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, Δ𝐻rxn .

Ca(s)+2H+(aq)⟶Ca2+(aq)+H2(g)

Answer 1

Looking at the balanced equation for the reaction taking place:

Ca(s) + 2HCl(aq) ==> CaCl₂(aq) + H₂(g) ... balanced molecular equation

q = mC∆T

q = ?

m = mass = 150.0 ml x 1.00 g / ml = 150. g

C = specific heat = 4.184 J/gº

∆T = change in temperature = +13.0º

q = (150. g)(4.1184 J/gº)(13.0º)

q = 8159 J

This is the heat generated by 0.601 g Ca. To find ∆Hrxn, we want the heat generated by 1 mol of Ca.

moles of Ca = 0.601 g x 1 mol Ca / 40.1 g = 0.01499 moles

∆Hrxn = 8159 J / 0.01499 moles = 544,296 J / mol = 544 kJ / mol (3 sig. figs.)