You carry out a reaction with 28.9mL of 0.711M solution of NiNO3)2. The reaction has a change in H of -26.4KJ/mol rxn. What is the q, the heat of the reaction, in joules?

I first found the mols of the solution using molarity and then found the joules of the reaction through conversion factors but I got the wrong answer trying to find joules for that amount of mols.

Any help is appreciated, thanks!

To find the heat of the reaction in joules (q), you need to use the equation:

q = ΔH × n

where ΔH is the change in enthalpy (in kJ/mol rxn) and n is the number of moles of the reactant (in mol).

Here's how you can calculate it step by step:

Step 1: Calculate the number of moles of Ni(NO3)2.

To do this, you can use the following equation:

n = M × V

where M is the molarity (in mol/L) and V is the volume (in L).

Given:
M = 0.711 M
V = 28.9 mL = 0.0289 L

n = 0.711 mol/L × 0.0289 L
n ≈ 0.0206 mol

Step 2: Convert the change in enthalpy from kJ/mol rxn to J/mol rxn.

Since the given ΔH is in kJ/mol rxn, you need to convert it to J/mol rxn by multiplying it by 1000:

ΔH = -26.4 kJ/mol rxn × 1000
ΔH ≈ -26,400 J/mol rxn

Step 3: Calculate the value of q, the heat of the reaction.

To find q, multiply the change in enthalpy (in J/mol rxn) by the number of moles of the reactant (in mol):

q = ΔH × n
q = -26,400 J/mol rxn × 0.0206 mol
q ≈ -543.84 J

Note that the negative sign indicates that the reaction is exothermic, meaning it releases heat.

Therefore, the heat of the reaction (q) is approximately -543.84 J.