In the calorimetry lab, the reaction mixture will consist of 2.00 mL 0.550 M HCl

Question

In the calorimetry lab, the reaction mixture will consist of 2.00 mL 0.550 M HCl(aq), 2.00 mL distilled water and 1.00 mL 1.00 M NaOH. Based on the following data, calculate the items below, each to three significant figures. The applicable equations are: q(reaction) + s(water) times mass(solution) times Delta T(solution) + 2.62J. degree C^-1 Delta T(water) = 0 Delta H(reactions, per H_2O formed) = q(reaction)/mols OH^- = q(reaction)/[NaOH] times L(NaOH) i) mols H Cl ____ mols ii) mols NaOH ____ mols iii) mass of solution ____ g iv) Delta T(solution) ____ degree C v) enthalpy of neutralization ____ kJ/mol

Explanation / Answer

Moles of HcL by using the above data 2.00 ml in 0.550 M

therefore , molarity = NUMBER OF MOLES / VOLUME OF SOLUTION IN L .

0.550 = (no. of moles/ 2.00 ml )* 1000

NO .OF MOLES = 0.0011

MOLES OF NaOh = MOLARITY * VOLUME

= 1* 1 /1000

0.001 MOLES OF Naoh

Mass of solution = (calorimeter + stir bar + solution )-( calorimeter + stirbar)

= 9.696- 4.617

= 5.079 g

TEMPERATURE = T final - T initail

= 22.45- 20.07

= 2.38

NaoH IN liters = 0.001 L

Enthalpy = 4.184 / 0.001 * 0.001 L

= 4,184000 J /MOL

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