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Calorimetry. Specific Heat of a Metal and Heat of Reaction 4. When 25.0 mL of 1.00 M sodium hydroxide, NaOH(aq) is added to 2
p4 VS P 77 pony PRELAB Analysis of Alka-Seltzer via Gas Evolution NAME DATE SECTION Pre-Laboratory Assignment 1. Complete and
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Answer #1

a) For question 4, the balanced reaction between NaOH and H2SO4 will be

2 NaOH (aq) + H2SO4 (aq) \rightarrow Na2SO4 (aq) + 2 H2O (l)

b) The qcal will be calculated by formula

qcal = Ccal . \Delta T

The \Delta T for reaction is given by formula \Delta T = (Tfinal - Tinitial)

\DeltaT = (31.2 - 22.3) = 8.9 oC.

Therefore qcal = 21.0 J / oC * 8.9 oC = 186.9 J = 187 J

Since qsolution = m.c.\DeltaT

Since mass of solution will be find by formula density = mass / volume.

Mass of solution = 25 + 25 = 50 ml

1.00 g / ml = m / 50 ml

Multiply both side by 50 ml.

Therefore 1.00 g / ml * 50 ml = m

m = 50g

qsolution = 50 g * 4.184 J / g oC * 8.9 oC = 1861.88 J = 1860 J

c) Since q for reaction will be -1861.88 J because heat absorbed by solution will be equal to heat lost by reaction and this heat lost will be negative. Therefore heat of reaction (qreaction) is -1861.88 J or -1860 J.

d) Since \Delta Hreaction = (qreaction / number of moles)

To find \Delta Hreaction first we need to find moles of acid by using formula molarity = moles / volume in litre.

Volume of H2SO4 = 25 ml * (1L / 1000 ml) = 0.025 L

Therefore rearranging formula of molarity for moles

moles = Molarity * volume in liter. Since M = moles / L. Therefore

moles = 1.00 moles / L * 0.025 L = 0.025 mol H2SO4.

For 0.025 mol we have qreaction -1861.88 J. For finding out for one mole we need to divide this value by 0.025.

Therefore, \Delta Hreaction = -1861.88 J / 0.025 mol = -74475.2 J / mol

By using dimensional analysis

-74475.2 J / mol * (1kJ / 1000 J) = -74.4752 kJ / mol.

After rounding off the answer will be -74.5 kJ / mol.

In multiple question we have to solve only one question. If you find any mistake please mention in comment box.

Thanks.

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