help with chem please

Question

Potassium nitrate, KNO3KNO3, has a molar mass of 101.1101.1 g/mol. In a constant-pressure calorimeter, 44.844.8 g of KNO3KNO3 is dissolved in 261261 g of water at 23.00 °C23.00 °C.

KNO3(s)−→−−H2OK+(aq)+NO−3(aq)KNO3⁢(s)→H2OK+⁡(aq)+NO3−⁢(aq)

The temperature of the resulting solution decreases to 18.80 °C18.80 °C. Assume that the resulting solution has the same specific heat as water, 4.184 J/(g·°C)4.184 J/(g·°C), and that there is negligible heat loss to the surroundings.

How much heat was released by the solution?

What is the enthalpy of the reaction?

Answer 1

KNO3 ==> H2O + NO3- + K+

For this problem we will use q = mC∆T

q = heat = ?

m = mass = 261 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = -18.80º (negative since temperature decreased)

q = (261 g)(4.184 J/gº)(-18.80º)

q = -20530 J ... this is the enthalpy for 1 mole of KNO3 according to the above reaction

We need to find how many moles are in the 44.84 g and then adjust our answer.

moles KNO3 = 44.84 g x 1 mol / 101.1 g = 0.4435 mols

∆Hrxn = 0.4435 mol x -20530 J/mol = -9105 J = -9110 J (3 sig. figs.)