........HCl + NaOH ==> NaCl + H2O
mols HCl = 0.1 x 1.00 = 0.1
mols NaOH = 0.1 x 1.00 = 0.1
mols H2O produced = 0.1
q = mass x specific heat H2O x (Tfinal-Tinitial)
q = 200 x 4.184 x (29.8-23.0)
That gives q for 0.1 mol in J. Convert to kJ/mol if desired.
Suppose 100.0 mL of 1.00 M HCl and 100.0 mL of 1.00 M NaOH, both initially at 23.0°C, are mixed in a thermos flask. When the reaction is complete, the temperature is 29.8°C. Assuming that the solutions have the same heat capacity as pure water, compute the heat released (in kJ).
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