NaOH (aq) + HCI (aq) NaCl (aq) + H20 (1) Volume of 1.00M HCI (mL) 45.0 Volume of 1.00 M NaOH (mL) 50.0 Initial Temperature (C) 22.1 Final Temperature ("C) 31.6 (a) Calculate the moles of HCl used in the reaction. Moles of HCI =.045 mol %3D (b) Calculate the moles of NaOH used in the reaction. Moles NaOH= .05 mol (c) Which reactant is the Limiting Reactant, HCl or NaOH?

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NaOH (aq) + HCI (aq) NaCl (aq) + H20 (1)
Volume of 1.00M HCI (mL)
45.0
Volume of 1.00 M NAOH (mL)
50.0
Initial Temperature (C)
22.1
Final Temperature ("C)
31.6
(a) Calculate the moles of HCl used in the reaction.
Moles of HCI = .045
mol
%3D
(b) Calculate the moles of NaOH used in the reaction.
Moles NaOH=.05
mol
(c) Which reactant is the Limiting Reactant, HCl or NaOH?
Limiting Reactant =
HCI
(d) Is this acid-base neutralization reaction exothermic or endothermic?
exothermic
NOTE: DO Include the correct number of significant figures or you will be marked wrong and DO NOT
include units in your answer.
Transcribed Image Text:NaOH (aq) + HCI (aq) NaCl (aq) + H20 (1) Volume of 1.00M HCI (mL) 45.0 Volume of 1.00 M NAOH (mL) 50.0 Initial Temperature (C) 22.1 Final Temperature ("C) 31.6 (a) Calculate the moles of HCl used in the reaction. Moles of HCI = .045 mol %3D (b) Calculate the moles of NaOH used in the reaction. Moles NaOH=.05 mol (c) Which reactant is the Limiting Reactant, HCl or NaOH? Limiting Reactant = HCI (d) Is this acid-base neutralization reaction exothermic or endothermic? exothermic NOTE: DO Include the correct number of significant figures or you will be marked wrong and DO NOT include units in your answer.
Expert Solution
Step 1ll

Yes! your answers are correct.

[NaOH] = 1.0M

VNaOH. = 50ml

(A)nNaOH. = 50mlx1.0 M =50 mmol = 0.050 mole 

[HCl]  = 1.0M

VHCl =   45 ml

(B)nHCl = 45mlx1.0 M =45 mmol= 0.045 mol

 

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