Question

lease write legibly. Round all final answers to the proper number of significant figures. Suppose you carry out an experiment to measure for the following reaction Nule] + 11 ng]- $ 140a + 14 (1) You follow the procedure outlined in the lab manual and acquire the following data using 50.0 mL of 1.0 M NHOH and 50.0 mL of 1,0 M HCl (for a total mass of 1000 g) A Temperature of calorimeter and NH OH before mixing: 22.1°C B. Temperature of HCl solution before mixing 22.1°C C Temperature of mixture (after mising): The calorimeter was tester earlier and found to have a heat capacity of 102/K What is the temperature difference (AT in that results upon mixing the two solutions? What is the heat gained in Joules) by the solution Show your calls for credit How much heated was gained in Joules) by the calorimeter? Show your calculations for credit What are the total joules released by the reaction Show your tons for credit How many moles of HCl are in 500 ml of a 10 M Ha solution Show your cakes forced Record your TEM 1411-Enthalpy of Neutralization How many moles of water are formed when 50.0 ml of 1.0M NHOH reacts with 50.0 ml of 10 Record your MHCI Show your callations for credit How many joules of energy are released by this reaction expressed per mole of water formed? Show your calculations for credit Record your How many kilojoules are released per mole of water

Answer 1

(a.) Temperature of NH₄OH before mixing = 22.1 ^oC

temperature of HCl before mixing = 22.1 ^oC

Average temperature = (temperature of NH₄OH before mixing + temperature of HCl before mixing) / 2

Average temperature = (22.1 ^oC + 22.1 ^oC) / 2

Average temperature = 22.1 ^oC

$\Delta$T = (temperature after mixing) - (Average temperature)

$\Delta$T = (28.5 ^oC) - (22.1 ^oC)

$\Delta$T = 6.40 ^oC

(b.) Heat gained by solution = (mass of solution) * (specific heat of solution) * ($\Delta$T)

Heat gained by solution = (100.0 g) * (4.184 J/g.^oC) * (6.40 ^oC)

Heat gained by solution = 2.68 x 10³ J

(c.) Heat gained by calorimeter = (calorimeter constant) * ($\Delta$T)

Heat gained by calorimeter = (10.2 J/^oC) * (6.40 ^oC)

Heat gained by calorimeter = 65.3 J

(d.) Total joules released by reaction = -(Heat gained by solution + Heat gained by calorimeter)

Total joules released by reaction = -(2.68 x 10³ J + 65.3 J)

Total joules released by reaction = -2.74 x 10³ J

(e.) moles of HCl = (concentration of HCl) * (volume of HCl in Liter)

moles of HCl = (1.0 M) * (0.0500 L)

moles of HCl = 0.050 mol

(f.)

moles of NH₄OH = (concentration of NH₄OH) * (volume of NH₄OH in Liter)

moles of NH₄OH = (1.0 M) * (0.0500 L)

moles of NH₄OH = 0.050 mol

moles of water formed = moles of HCl = moles of NH₄OH = 0.050 mol

(g.) energy released = (Total joules released by reaction) / (moles of water formed)

energy released = (-2.74 x 10³ J) / (0.0050 mol)

energy released = -5.5 x 10⁴ J/mol H₂O

(h.) energy released = -5.5 x 10⁴ J/mol * (1 kJ / 1000 J)

energy released = (-5.5 x 10⁴ / 1000) kJ/mol H₂O

energy released = -55 kJ/mol H₂O