A student followed the procedure in this experiment, using 50.00 mL of 1.00 M HC

Question

A student followed the procedure in this experiment, using 50.00 mL of 1.00 M HCl and 51.00 mL of 1.00 M NaOH both solutions having an initial temperature of 25.5 oC. After mixing, a maximum temperature of 32.2 oC is measured. For this problem, assume the density of the solution is 1.040 g/mL and its specific heat is 4.18 J/g oC. Report all answers with the proper number of significant figures. Units are designated in the problem. Do NOT use scientific notation for answers.

Based on this information, the number of moles of water formed is:

moles.

The temperature change observed is

oC.

The mass of solution is

grams.

Based on all these values, the heat transferred in this reaction is

Joules.

This yields a Heat of Neutralization of

kJ/mol.

Explanation / Answer

total mass = 50 + 51 = 101 mL = 101*1.04 = 105.04 g

HCl + NaOH = H2O + NaCl

dT = 3

32.2-25.5 = 6.7°C

Q = m*C*(dT) = 105.04*4.184*6.7 =2944.565 J

for..

HRxn = -Q/n = -2944.565 /(50*1) = 58.8913 kJ/mol

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