s H230c How many kJ of heat are released when 85.0 g of NaOH reacts with Na2SO4(

Question

s H230c How many kJ of heat are released when 85.0 g of NaOH reacts with Na2SO4(aq) + 2 H2O() +1? 4. HaSO4(aq) + 2 NaOH(aq) ? 5. 2.50 g of octane, Ceths was burned in a bomb calorimeter causing the temperature to increase from 21.0°C to 35.5°C. The total heat capacity of the calorimeter was 8.33 kJPC. Calculate the heat of combustion for the reaction shown below. 2CsHs(l) + 2502(g) ? 16 CO2(g) + 18H20(g) 6. When 50.0 mL of 0.100 M AgNOs was reacted with 50.0 mL of 0.100 M HCl in a Styrofoam cup calorimeter, the temperature increased from 22.6°C to 23.4°C. Calculate AH for the following reaction assuming the density and specific heat of the solutions are 1.00 g/mL and 4.18 JigoC respectively. AgNOs(aq) HCl(aq) AgCI(s) + HNO (aq)

Explanation / Answer

1) when two moles of NaOH reacts with one mole of sulfuric acid, 114 kJ of energy is released

Now as we have taken excess of H2SO4, so the limiting reagent is NaOH

Mass of NaOH taken = 85g

Moles of NaOH taken = Mass / molar mass = 85 / 40 = 2.125 moles

So the heat released = 114 X 2.125 / 2 = 121.125 kJ

8) the heat evolved during the combustion is absorbed by calorimeter

Heat absorbed = Heat capacity X change in temperature = 8.33 kJ /0C X (35.5-21) = 120.79 kJ

This heat is evolved when 2.5g of octane is burnt

in the given reaction two moles of octane are undergoing combustion

Molar mass of octane = 114.23 g / mole

Mass of octane in the reaction = 2 X 114.23 grams = 228.46g

Heat evolved by burning one gram = 120.79 / 2.5 kJ = 48.32 kJ

Heat evolved by burning of 228.46g = 11038.27 kJ

9) total volume of solution = 50 + 50 = 100 mL

the mass of solution = 100g (as density is 1g/mL)

Heat evolved when 50mL each of 0.1 M concentration AgNO3 and 0.1 M conc of HCl reacts

= mass of water X specific heat X change in temperature = 100 X 4.18 X (23.4-22.6) = 334.4 Joules

Number of moles reacted = Molarity X volume in L = 0.1 X 50 / 1000 = 0.005 moles

So the heat evolved per mole = 334.4 / 0.005 = 66880

Or enthalpy change = -66880 J / mole = -66.88 kJ / mole

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