A 98.2 mL sample of 1.00 M NaOH is mixed with 49.1 mL of 1.00 M H2SO4 in a large

Question

A 98.2 mL sample of 1.00 M NaOH is mixed with 49.1 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 22.25 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 30.30 °C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g·°C), and that no heat is lost to the surroundings.

1st attempt

Part 1(1 pt)

Part 2(1 pt)

Part 3(1 pt)

Explanation / Answer

Part 1 :
2 NaOH + H2SO4 -----> Na2SO4 + 2 H2O
Part 2:
mol of NaOH = (98.2/1000)L * 1 mol/L = 0.0982 mol
mol of H2SO4 = (49.1/1000)L * 1mol/L = 0.0491 mol
According to the stoichiometry,
mol of  H2SO4 reacted with NaOH = 0.0982mol of NaOH*(1mol H2SO4 / 2 mol NaOH) = 0.0491 mol
Hence no reactants are left behind after the reaction. So the answer is option A (No)

Part 3 :
mass = (98.2mL+49.1mL) * 1g/mL = 147.3 gm
Heat gained = -mCp(T2-T1) = 147.3gm * 4.18 J/gmoC*(30.30-22.25)OC = -4956.4977 J
enthalpy change per mole of H2SO4= - (4956.4977/1000)kJ / 0.0491mol = -100.947 kJ/mol

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