When 150.0 mL of a 0.500 M citric acid (H_3C_5H_5O_7) solution reacts with 50.0

Question

When 150.0 mL of a 0.500 M citric acid (H_3C_5H_5O_7) solution reacts with 50.0 mL of a 3.00 M KOH solution in a calorimeter, the temperature of all components rises from 21.2 degree C to 28.7 degree C. Assume the density and heat capacity of the solutions are the same as that of water. H_3C_6H_5O_7(aq) + 3 KOH(aq) rightarrow K_3C_6H_5O_7(aq) + 3 H_2O(l) Is there a limiting reagent? If so, what is it? Calculate DeltaH_rxn for the reaction as written. (Once you discover that KOH is the limiting reagent has to be divided by three.)

Explanation / Answer

The reaction between citric acid and KOH can be represented as

C6H8O7 + 3 KOH = C6H5K3O7 + 3 H2O

stoichiometric ratio of reactants citric acid and KOH= 1:3

The reaction suggests 1 mole of Citric acid reacts with 3 molesof KOH.

given that 150 ml of 0.5 M citric acid

moles of Citric acid in 150 ml =(150/1000)*0.5=-075 moles

moles of KOH in 50 ml of 3M KOH = (50/1000)*3= 0.15 moles

moles of Citric acid reacted =0.15/3 =0.05

Stoichiometric ratio of reactants taken : 0.075: 0.15 = 1:2

This suggests that KOH is the limiting reactant

b)

moles of citric acid reactged =0.05 moles

molecular weight of citric acid( (C6H8O7)= 12*6+ 8*1+16*7= 192

moles of citric acid =192*0.05=9.6 gms

delH for citric acid= mass * heat capacity* temperature difference = 9.6 * 4.18 (j/g.deg.c)*(28.7-21.2)=300.96 joules

moles of KOH= 0.15 moles

its moleculr weight (KOH)= 39+16+1= 56

mass of KOH= 0.15*56=8.4 gms

delH =8.4*4.18*(28.7-21.2)=263.34 joules

total delH= 300.96+263.34= 564.3 Joules

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