Verifying Calculations for 2g of NH4Cl: Low Yields Explained

[mastiffcacher]

Trying to make sure my calculations were correct. We were tasked to make 2 grams of NH₄Cl. Our proposed method was using HCl and NH₄ClOH. This was confirmed as our best way.

The way I figured it was that 2 grams of ammonium chloride was 0.037519 moles. We used 6M hydrochloric acid and 6M ammonium hydroxide for the reaction. I calculated that we needed 6.25 mL of each solution to make the 2 grams. Our results were horrible as we only made .186 g and .233 g in two different trials.

I am wondering if these amounts are correct at least theoretically. Does this reaction typically have this low of a percent yield? If it is usually this low of a yield, why might this be?

 

[chemisttree]

When you try to dry ammonium chloride, it spontaneously reverts back to HCl and NH₃ under suprisingly mild conditions... it evaporates.

 

[mastiffcacher]

So it should have produced 2 grams but the loss is great while drying?

 

[chemisttree]

What conditions did you use to produce and dry it?

 

[mastiffcacher]

We used 6.25 mL of 6M HCl and 6.25 mL of 6M NH4OH. Placed in beaker under hood. When vapors stopped from beaker, we moved to a hot plate se at ~250F. It took about 15 minutes or so for the resultant water to evaporate. This was done in my chem lab for school. I was working on a report and am trying to figure out what errors may have occurred. I just wanted to make sure that I did not miss something in my calculations and that 6.25 mL was the correct volume of 6M reactants to use. I also thought that my TA told us that this was the better reaction to run. Maybe she meant the easiest. If that is the correct volume, why is the loss so great?

 

[Borek]

Calculations are OK. The only thing I can think off - apart from some human error - is what was already signalled by Chemisttree, loss due to the sublimation of the solid.