REPORT ON PART II: ANALYSIS OF AMMONIA Trial 1 Standardization of NaOH: Trial 2

Question

REPORT ON PART II: ANALYSIS OF AMMONIA Trial 1 Standardization of NaOH: Trial 2 Molarity of Standard HCI Final buret reading for Ha Initial buret reading for HCI Volume of Ha Final buret reading for NaoH Initial buret reading for NaoH -21ml- Volume of NaOH Molarity of NaOH solution Average molarity Trial 1 Trial 2 Analysis of Ammonia: Mass of flask and sample Mass of flask Mass of sample Molarity of HCI Final buret reading of HCI Initial buret reading of HCI Volume of HCl Moles of HCI added Molarity of NaOH solution Final buret reading of NaoH 9.4 Initial buret reading of NaoH volume of NaoH Moles of NaOH added Moles of HCl consumed by compound Moles of NHh in sample Moles of NH) per gram of sample Average moles of NHh per gram of sample Mass percentage of NHh Trial 3 Trial 3 2.00

Explanation / Answer

Consider the readings for trial 2. The following calculations will make the concept clear.

Moles of HCl added = Volume used in liters*Molarity = 0.012*0.312 = 0.003744 moles

Similarly,

Moles of NaOH added = Volume used in liters*Molarity = ((13.8-9.42)/1000)*0.150 = 0.000657 moles

So, moles of HCl present to react with NaOH = 0.000657

So, moles of HCl consumed by NH3 = 0.003744-0.000657 = 0.003087

Since 1 mole NH3 reacts with 1 mole HCl, so moles of NH3 present in sample = 0.003087

You have not provided the sample mass, so further calculations are not possible.

Hope this helps ! Revert back if you have any queries.

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