Chemistry, 08.10.2019 02:00 wtwbegay

Acommon laboratory reaction is the neutralization of an acid with a base. when 32.8 ml of 0.500 m hcl at 25.0°c is added to 66.3 ml of 0.500 m naoh at 25.0°c in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°c. what is the heat of reaction per mole of nacl (in kj/mol)? assume the mixture has a specific heat capacity of 4.18 j/(g·k) and that the densities of the reactant solutions are both 1.07 g/ml.

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Answer from: tyrianadonelson

Explanation:

The given data is as follows.

Densities of both the reactant solutions = 1.07 g/mL.

Specific heat = 4.18 $J/g^{o}C$

Volume of HCl = 32.8 ml

Volume of NaOH = 66.3 ml

The reaction will be as follows.

$HCl + NaOH \rightarrow NaCl + H_{2}O$

Since, density is mass divided by volume. Hence, calculate the mass of HCl as follows.

Mass of HCl = Density × Volume of HCl

= $1.07 \times 32.8 ml$

= 35.096 g

Similarly, mass of NaOH will be calculated as follows.

Mass of NaOH = Density × Volume of NaOH

= $1.07 g/ml \times 66.3 ml$

= 70.941 g

Therefore, total mass will be as follows.

Total mass = Mass of HCl + Mass of NaOH

= 35.096 g + 70.941 g

= 106.037 g

Change in temperature will be calculated as follows.

$\Delta T$ = $(28.2 - 25)^{o}C$

= $3.2^{o}C$

Therefore, calculate the heat of reaction as follows.

q = $mC \Delta T$

= $106.037 g \times 4.18 J/g^{o}C \times 3.2^{o}C$

= 1418.35 J

or, = 1.418 kJ (as 1 kJ = 1000 J)

No. of moles of HCl = Molarity × Volume

= 0.5 M × 32.8 ml

= 16.4 mol

No. of moles of NaOH = Molarity × Volume

= 0.5 M × 66.3 ml

= 33.15 mol

So, HCl is the limiting reagent and heat of reaction produces by per mole of NaCl will be calculated as follows.

Heat released for 1 mole of NaCl = $\frac{1.418 kJ}{16.4 mol}$

= 0.0864 kJ/mol

Thus, we can conclude that the heat of reaction per mole of NaCl is 0.0864 kJ/mol.

Answer from: Quest

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Acommon laboratory reaction is the neutralization of an acid with a base. when 32.8 ml of 0.500 m hc...

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