In a student's experiment, when 40 mL of 6% (w/w) hydrogen peroxide solution was reacted with 10 mL 0.5 M FeCl3 solution, a ΔrH of -3,382 J was calculated.

Calculate ΔH for this reaction in kJ mol-1. Give your answer to 2 decimal places, do not include the units in your answer. Assume a density of 1.0 g mL-1 for the hydrogen peroxide solution.

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6% solution means 6g H2O2 per 100g of water. 1g of water = 1ml, so 6g H202/100mL. Now you have 40mL solution, so, 40mL * (6g/100mL) = 2.4g of H2O2, moles of H2O2 = 2.4/34.0147 = 0.0705 moles H2O2. ΔH = ΔrH/moles H2O2.

ΔH = -3.382kJ/0.0705mole = 47.93 kJ/mol

-47.93*

To calculate ΔH for the reaction, we need to convert the given values to appropriate units and then use the formula:

ΔH = ΔrH / moles reacted

First, let's calculate the moles of hydrogen peroxide reacted by using the concentration and volume of the solution:

moles H2O2 = concentration * volume * density
= (6 g/100 g) * (40 mL/1000 mL) * (1 g/mL)
= 0.024 g

Next, convert the grams of hydrogen peroxide to moles using its molar mass. The molar mass of hydrogen peroxide (H2O2) is about 34.02 g/mol.

moles H2O2 = 0.024 g / 34.02 g/mol
≈ 0.000706 mol

Now, we can calculate ΔH:

ΔH = ΔrH / moles reacted
= -3,382 J / 0.000706 mol
≈ -4,791,500 J/mol

Finally, convert the answer to kJ/mol by dividing by 1000:

ΔH ≈ -4,791.50 kJ/mol

Therefore, ΔH for this reaction is approximately -4,791.50 kJ/mol.