Don't understand this calorimetry question in regards to the mols

8) 50 cm3 of 1.0 mol dm-3 nitric acid was added to 20 cm3 of 1.0 mol dm-3 barium hydroxide solution. The temperature rose by 7.9C. Calculate the enthalpy of neutralisation for this reaction (per mole of nitric acid reacting). Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.

So, what I thought I'd do was the following:

m = 70 deltaT = 7.9 c = 4.18

q = 70 * 7.9 * 4.18 = 2312J = -2.312kJ

Mols = conc * volume and we use it in regards to the mole of nitric acid reacting so...

50cm^3 = 0.05dm^3

0.05dm^3 * 1.0 mol/dm^3 = 0.05mols

So to find the kjmol-1, it would be -2.312kj/0.05mols, right? Well, everything else is right but the answers go 0.04 mols which confused me. Can someone explain this?

I think I've understood it. Someone correct me if I'm wrong...

If we calculate the mols of both reactants...

Nitric acid is at 0.05mols

Barium hydroxide solution would be 0.02dm^3 * 1 = 0.02mols.

Nitric acid is at excess so the only reactant fully reacting would be the 0.02mols of Barium hydroxide solution. However, since we have to calculate it per mole of nitric acid reacting, we have to find the mol of nitric acid that reacted, so we write out the equation...

2HNO3 + Ba(OH)2 ---> Ba(NO3)2 + 2H2O

Seeing as there's 2 lots, then there's twice as much mols, making 0.04 thus where they got 0.04.

I've understood that but in the exam, I'd never be able to write out the equation for the solution correctly. Slightly worrying...

I know I’m late but can someone tell me how they got the mass of 70

I think I've understood it. Someone correct me if I'm wrong...
If we calculate the mols of both reactants...
Nitric acid is at 0.05mols
Barium hydroxide solution would be 0.02dm^3 * 1 = 0.02mols.
Nitric acid is at excess so the only reactant fully reacting would be the 0.02mols of Barium hydroxide solution. However, since we have to calculate it per mole of nitric acid reacting, we have to find the mol of nitric acid that reacted, so we write out the equation...
2HNO3 + Ba(OH)2 ---> Ba(NO3)2 + 2H2O
Seeing as there's 2 lots, then there's twice as much mols, making 0.04 thus where they got 0.04.
I've understood that but in the exam, I'd never be able to write out the equation for the solution correctly. Slightly worrying...

God bless you man this explanation is the best