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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.

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  • siggy
    Lv 6
    9 years ago
    Favorite Answer

    Formula is -

    NaOH + HNO3 => H2O + NaNO3

    moles of base = moles acid = CV = 0.300 x 0.100 = 0.03 moles

    Heat produced = ∆ H = mC∆T = 200 x 4.18 x 2 = 1672 Joules

    where m = mass of water, C = heat capacity water = 4.18 J/g/K

    ∆ T = temperature change of water

    Since you get 1672 J per 0.03 moles reactant, then -

    the heat evolution for 1 mole would be 1672 x 1 / 0.03 = 55730 J = 55.73 kj/mol

    That's close - the actual figure should be 57.1 kJ/mol. .

    Source(s): chem background
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