Question

Trial #1	CaCl2	KNO3
Mass (g)	1.16 g	1.22 g
diH2O (mL)	10.0 mL	10.0 mL
Initial Temperature (T ioC)	22.8 oC	23.6 oC
Final Temperature (TfoC)	34.4 oC	15.7 oC
Trial #2
Mass (g)	1.13 g	1.15 g
diH2O (mL)	10.0 mL	10.0 mL
Initial Temperature (TioC)	23.3 oC	23.8 oC
Final Temperature (TfoC)	33.8 oC	15.5 oC

Calculate heat (qrxn) for each trial using q = (m)(s)(Δ). Assume that the specific heat capacity of the salt solution is the same as water (4.184 J/g∙°C)4.184 J/g∙°C , then calculate the moles of the salt for each trial.

Calculate ∆Hrxn for each trial and then calculate the average ∆Hrxn for each salt.

Create a data analysis table for the results. The table should have each trial in a new column. Row values should include ∆T, q, n, ∆Hrxn, average ∆Hrxn.

Calculate % error by comparing the calculated value to the actual value.

Answer 1

The balanced reaction for the given salt solution is as follows.

CaCl₂ + 2KNO₃ Ca(NO₃)₂ + 2KCl

Trial #1

The no. of moles of CaCl₂ = 1.16 g / 111 g mol^-1 = 0.010 mol

The no. of moles of KNO₃ = 1.22 g / 101.1 g mol^-1 = 0.012 mol

Therefore, the no. of moles of Ca(NO₃)₂ = 0.012/2 = 0.006 mol (= n)

Note: Here, KNO₃ is the limiting reagent, which reacts with half of its no. of moles CaCl₂ to form the same no. of moles of Ca(NO₃)₂.

The mass of Ca(NO₃)₂ = 0.006 mol * 164.086 g mol^-1 = 0.99 g

Now, q_rxn = 0.99 g * 4.184 J/g∙°C * {(34.4+15.7)/2 - (22.8+23.6)/2} °C

= 7.663 J

Now, ∆H_rxn = q_rxn/n

= 7.663 J /0.006 mol

= 1.270 KJ/mol (1 KJ = 10³ J)

Trial #2

The no. of moles of CaCl₂ = 1.13 g / 111 g mol^-1 = 0.010 mol

The no. of moles of KNO₃ = 1.15 g / 101.1 g mol^-1 = 0.0114 mol

Therefore, the no. of moles of Ca(NO₃)₂ = 0.011/2 = 0.0057 mol (= n)

Note: Here, KNO₃ is the limiting reagent, which reacts with half of its no. of moles CaCl₂ to form the same no. of moles of Ca(NO₃)₂.

The mass of Ca(NO₃)₂ = 0.0057 mol * 164.086 g mol^-1 = 0.93 g

Now, q_rxn = 0.93 g * 4.184 J/g∙°C * {(33.8 + 15.5)/2 - (23.3 + 23.8)/2} °C

= 4.295 J

Now, ∆H_rxn = q_rxn/n

= 4.295 J /0.0057 mol

= 0.755 KJ/mol (1 KJ = 10³ J)

Now, the average ∆H_rxn for both the trials = (1.270 + 0.755)/2 = 1.013 KJ/mol