[Win,odd Dhamnekar]

The addition of 3.15 g of Ba(OH)₂·8H₂O to a solution of 1.52 g of NH₄SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:
Ba(OH)₂·8H₂O(s) + 2NH₄SCN(aq) Ba(SCN)₂(aq) + 2NH₃(aq) + 10H₂O(l)

Calculate ΔH for the reaction described by the above equation.

My answer: q=cmΔT
                   =4.20 J/g°C × (3.15 + 1.52 +100)g ×3.1°C
                   =1362.8 J =1.4 kJ (two significant figures)
 ΔH for the reaction :
We have 3.15 g × [itex]\frac{1 mol}{315.46 g} =0.00998 [/itex]mol Barium Hydroxide Octahydrate available. 1.52 × [itex]\frac{1 mol}{76.1209 g} = 0.02[/itex] mol ammonium thiocynate available.

Since 0.02 mol of NH₄SCN × [itex]\frac{1 mol Ba(OH)_2\cdot 8H_2O}{2 mol NH_4SCN}=0.01  mol Ba(OH)_2 \cdot 8H_2O [/itex]is needed. Ba(OH)₂·8H₂O is limiting reagent.

The reaction uses 2 mol NH₄SCN and the conversion factor is [itex]\frac{1.4kJ}{0.02 mol NH_4SCN}[/itex], so, we have
[tex] \Delta {H}= 2 mol \times \frac{1.4 kJ}{0.02 mol NH_4SCN}=+6700 J [/tex]

The enthalpy change for this reaction is +6700 J and thermochemical equation is
  Ba(OH)₂·8H₂O + 2NH₄SCN  Ba(SCN)₂ + 2NH₃ + 10H₂O ΔH = +6.7 kJ

 Is this correct?
In my opinion, it looks correct. 

[mjc123]

How do you calculate 2*1.4/0.02 = 6.7???

A couple of smaller points:
(i) If you calculate an intermediate result, use the more precise form in subsequent calculations, and round off at the end. Repeatedly rounding off can lead to significant errors. If you are asked to give the intermediate result as an answer, you can round off the answer (e.g. 1.4 kJ), but still use the more precise form (1363 J) for subsequent calculations.
(ii) You did right to determine the limiting reagent, though in this case 0.00998 is so close to 0.01 that the mixture may be regarded as stoichiometric. But if one reagent is limiting, that is the one you should use in the next step.

[Win,odd Dhamnekar]

 I think my computed answer is wrong. [tex]\Delta{H}= 2mol \times \frac{1363 J}{0.02 mol NH_4SCN}= 136300 J =136.3 kJ [/tex] is correct.

[mjc123]

OK, so if you were rounding the final answer to 2 sig figs...?

[Win,odd Dhamnekar]

 If i were to round off the final answer up to two significant figures, it would be 136.30 kJ.

[Borek]

If i were to round off the final answer up to two significant figures, it would be 136.30 kJ.

That's five sigfigs.

[Win,odd Dhamnekar]

Then 1.4 kJ must be correct.

[Borek]

Then 1.4 kJ must be correct.

No, it is not.

Apart from the correct notation and correct application of sig figs 136.30 kJ is almost 100 times more than 1.4 kJ.

[Win,odd Dhamnekar]

 Then 1.400 kJ is correct. isn't it?

[Borek]

Instead of guessing at random and wasting time you should follow simple rules that tell how to use significant figures.

No, 1.400 kJ is not correct, it is still 100 times lower than the value you were told is a correct one.

[Win,odd Dhamnekar]

1.4 × 10³ kJ Why are significant figures so important in chemistry? in mathematics and statistics it is not so important.

[Borek]

No.

I am locking the thread, we are not going to play your guessing game.