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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

Answer from: Quest

The reaction will shift to the right

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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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