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

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lemonade12345
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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?
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lemonade12345
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Anyone?
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lemonade12345
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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...
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charco
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(Original post by lemonade12345)
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?
Mass = 0.07 kg

Moles of nitric acid available = 0.05
Moles of barium hydroxide available = 0.02

BUT

2 mol nitric acid reacts with 1 mol barium hydroxide

Therefore 0.02 mol barium hydroxide reacts with 0.04 mol nitric acid

ΔE = 0.07 x 4.18 x 7.9 = 2.31 kJ

per mol = 2.31/0.04 = -57.8 kJ
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gbriett
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how did you figure out the mass?
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Adhamsan
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(Original post by lemonade12345)
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...
Thank you!!!!!!!!!!!!!
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