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lexas Sta ning comibiscms/mod/ibis/view php?id=4189275 Jump to.. Map d Sapling Learning In a constant-pressure calorimeter, 70.0 mL of 0.750 M H2SO, was added to 70.0 mL of 0.290 M NaOH The reaction caused the temperature of the solution to rise from 24.22 to 26.20 . If the solution has the same density and specific heat as water (1.00 g/ml. and 4.184 Jig K, respectively), what is H for this reaction (per mole of H20 produced)? Assume that the total volume is the sum of the individual volumes. Number kJ/ mol H,0 0 Check Answer 0 Next Exit O Pre vous Grve Up & View Solution Hint rch

Explanation / Answer

Total volume of the solution = 70.0 + 70.0 = 140. mL

Mass of the solution = density * volume = 1.00 * 140. = 140. g.

Specific heat of the solution = 4.184 J

Change in temperature = 26.20 - 24.22 = 1.98 0C

So,

Change in heat = mass * specific heat * change in temperature

q = 140. * 4.184 * 1.98

q = 1160 J

Balanced equation is,

H2SO4 (aq.) + 2 NaOH (aq.) -------------> Na2SO4 (aq.) + 2 H2O (l)

Initial moles of H2SO4 = 0.750 * 70.0 / 1000 = 0.0525 mol

Initial moles of NaOH = 0.290 * 70.0 / 1000 = 0.0203 mol

From the balanced equation,

1 mol of H2SO4 needs 2 mol of NaOH

Then,

0.0525 mol of H2SO4 needs 2 * 0.0525 = 0.105 mol of NaOH

But we have only 0.0203 M NaOH. Hence, NaOH is limiting reagent. It decides the extent of reaction.

From the balanced equation,

2 mol NaOH forms 2 mol H2O

Then, 0.0203 mol of NaOH forms 0.0203 mol of H2O

Therefore,

deltaH = q / n = 1160 / 0.0203 = 57143 J = 57.1 kJ

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