Mole product if HCl limiting Mole product if NaOH limiting Limiting reactant Del
ID: 1051756 • Letter: M
Question
Mole product if HCl limiting Mole product if NaOH limiting Limiting reactant Delta H (neutralization) Accepted Delta H Percent error Part 2: Hneutralization Trial 1 Trial 2 6227 Mass of 1.00 MHCI 62.120 9 Vol. of 1.00 M HCI e n1(elcealated) 62,827mL _(ealculatcd g Vol. of 1.00 M HCl had 20 m 2 (calculated) Vol. of 1.00 M NaOH 5mL Mass of 1.00 M NaOH 50 0 ( Mole product if 827(ealculated) 55,i m (calculated) (calculated) 55 (calculated) HCl limiting reactant (calculated) (calculated) Mole product if NaOH limiting reactant (calculated) (calculated)Explanation / Answer
Trial 1
Trail 2
Mass of 1.0 M HCl (g)
62.120
62.827
Volume of 1.0 M HCl (mL)
62.120
62.827
Vol. of 1.0 M NaOH (mL)
50.0
55.1
Mass of 1.0 M NaOH (g)
50.0
55.1
Mole product if HCl limiting reactant **1
0.06212
0.062827
Mole product if NaOH limiting reactant **2
0.050
0.0551
**1 and 2 The neutralization reaction is
HCl (aq) + NaOH (aq) ------> NaCl (aq) + H2O (l)
There is a 1:1 molar reaction between HCl and NaOH.
Moles HCl added = (vol. of HCl in L)*(concentration of HCl in moles/L) = (62.120 mL)*(1 L/1000 mL)*(1.0 mole/L) = 0.06212 mole.
Moles NaOH added = (vol. of NaOH in L)*(concentration of NaOH in moles/L) = (50.0 mL)*(1 L/1000 mL)*(1.0 mole/L) = 0.050 mole.
Since we are working out the heat of neutralization per mole of water, therefore, water (H2O) is our desired product.
There is a 1:1 molar ratio between HCl and H2O as well as 1:1 molar ration between NaOH and H2O.
Therefore, when HCl is the limiting reactant,
moles H2O produced = (0.06212 mole HCl)*(1 mole H2O/1 mole HCl) = 0.06212 mole.
When NaOH is the limiting reactant,
moles H2O produced = (0.050 mole NaOH)*(1 mole H2O/1 mole NaOH) = 0.050 mole.
In both the cases, we see that NaOH produces a smaller amount of the product and therefore, NaOH is the limiting reactant here.
Therefore,
Trial 1
Trial 2
Limiting reactant
NaOH
NaOH
Moles H2O produced
0.050
0.0551
Trail 1
Trial 2
Total mass in calorimeter (g)
112.12
117.927
Initial temperature NaOH (C)
28.9
22.0
Initial temperature HCl (C)
22.1
22.0
Average initial temperature (C)
25.5
22.0
Final temperature (C)
29.5
28.9
Change in temperature (C)
4.0
6.9
Hreaction = (total mass in calorimeter)*(4.184 J/g.C)*(change in temperature) (J)
1876.440
3404.505
Hreaction (in kJ)
1.876
3.404
Hneutralization = -Hreaction/(moles of water produced) (kJ/mol)
-37.52
-61.778
Average Hneutralization (kJ/mol)
(-37.52)+(-61.778)/2 = -49.649
Accepted Hneutralization (kJ/mol)
-55.2 (obtained from internet sources)
Percent error
10.056
Trial 1
Trail 2
Mass of 1.0 M HCl (g)
62.120
62.827
Volume of 1.0 M HCl (mL)
62.120
62.827
Vol. of 1.0 M NaOH (mL)
50.0
55.1
Mass of 1.0 M NaOH (g)
50.0
55.1
Mole product if HCl limiting reactant **1
0.06212
0.062827
Mole product if NaOH limiting reactant **2
0.050
0.0551
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