Question

Concentration of NaOH = 1.1M

Volume of NaOH = 45mL

Concentration of HCL = 1.2M

volume of HCL = 45mL

final temperature of mixture = 29.1 degrees celsius

initial temperature (HCL and NaOH) = 21.9 degrees celsius

mass of final mixture (assume density of mizture is 1.00g/mol) = 90.0g

specific heat capacity for H20 = 4.18J/g/degree celsius

1) calculate qwater, qcalorimeter and qreaction

2) calculate the number of mols reacted of the limiting reagent

Answer 1

Delta H of neutralization reaction is -55840 J/mol

Determine limiting reactant.

Mol NaOH = volume in L x molarity = 0.045 L x 1.1 mol /L = 0.0495 mol

Mol HCl = 0.045 L x 1.2 mol/L=0.054

Since mol ratio is 1:1 , NaOH is limiting reagent.

Answer to second part (2) is moles of limiting reactant reacted is 0.0495 mol

Calculate qrxn using limiting reactant

q rxn = delta H x n

n is moles of limiting reactant

= -55840 x 0.0495 = -2764.08 J

Calculate Kcal

-qrxn = qwater + qcal

-(-2764.08) = 90.0 g x 4.18 J/g deg C x (29.1-21.9) + Kcal x (29.1 – 21.9)

Kcal = 7.7

qCal = Kcal x delta T= 7.7 J / deg C x (29.1-21.9) = 55.44 J

q water = 90.0 x 4.18 x (29.1 -21.9) = 2708.64 J

answer 1)

q water = 2708.64 J

q rxn = -2764.08 J

q cal = 55.44 J

2)

Number of moles of reagent reacted = 0.0495 mol