Question

100.0 mL of 2.000 M sodium hydroxide is mixed with 200.0 mL of 1.000 M nitric acid. Both solutions are initially at a temperature of 23.00 °C, the resulting sodium nitrate solution has a temperature of 31.91 °C What is the heat of neutralization (H) for this reaction? [Assume that the density and heat capacity of all solutions are identical to that of water: 1.000 g/mL and 1.000 cal/g°C and that there is no loss of heat from the solution.]

Answer 1

What reaction? first write the reaction

Answer 2

Then use: \(q=m*C_p*\Delta T\)

and \(\Delta H^o=\dfrac{q}{n}\)

q= heat, m=mass, \(\Delta T\)= change in temp, n= moles

Answer 3

@aaronq i have the reaction written out but how do i find the mass... which mass do i use for that equation?

Answer 4

use the mass of the product

Answer 5

use the mass of the salt in the product convert to moles and divide by the stoichiometric coefficient (i'm pretty sure, since you have to normalize for the reaction)