In my organic chemistry textbook, lithium is used to create a free radical of the alkynes, to then allow the anti addition of hydrogen to get an alkene.
"The Dissolving Metal Reduction of Alkynes" is a reaction that then uses $\ce{NH4Cl}$. Why put $\ce{NH4Cl}$?
Answer
Products from reaction step (1) are the trans-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.
$$\ce{LiNHEt + NH4Cl ->~ EtNH2 + NH3 + LiCl}$$ $$\ce{Li + H2O ->~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$
Secondly, the trans-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.
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