NaOH + HCl ==> NaCl + H2O
mols NaOH = 0.100 L x M = ?. We need to find the M and we can do that if know the mols.
q for neutralization is 55.90 kJ/mol. We had q of -2.676 kJ so # mols in the reaction must have been -2.676/-55.90 = 0.04787
M NaOH = mols/L = 0.04787/0.1L = ? mols/L.
The following solutions are combined in a calorimeter to determine the enthalpy of the neutralization reaction.
100.0 mL sodium hydroxide, NaOH, of unknown concentration an excess of 1.3732 M hydrochloric acid, HCl.
The experimental value for q is determined to be -2.676 kJ
Knowing that the theoretical value of q for any neutralization reaction is -55.90 kJ/mol H2O, determine the concentration of the sodium hydroxide solution.
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