Question

0.028 g of finely powered Ca(s) are added to 10.00 mL of 1.042 M HCl in a well of a Styrofoam block calorimeter with K_cal = 6.8 J/^oC. The temperature of the acid solution increases from 24.2 ^oC to 32.5 ^oC. Calculate ΔH for the reaction that occurred.

Answer 1

the reaction

Ca(s) + 2HCl(aq) --> CaCl2 + H2(g)

so..

MW of Ca = 40 g/mol

mol of Ca = mass/MW = 0.028/40 = 0.0007 mol of Ca

mol of HCl = MV = (10*10^-3)(*1.042) = 0.01042 mol of HCl

so..

ratio is 1:2

0.0007 mol of CA --> 2*0.0007 = 0.0014 mol of HCl reqruied, so we have enough and calcium is limiting reaction

so..

Qcalorimteter = C*dT = (6.8)*(32.5-24.2) = 56.44 J

then... assume

Qlost = -56.44 J

for

HRxn = Qlost / n = -56.44 / 0.0007 = -80628.571J/mol = -80.629 kJ/mol