Complete the reaction: #HCN + 2H_2O -> ?# Also, how is oxamide prepared? What is the reaction for its preparation?

Nitriles undergo hydrolysis in acid on high heat to form carboxylic acids. Thus, hydrocyanic acid hydrolyzes into formic acid.

The full mechanism for a general nitrile is:

1. The nitrile is protonated on the nitrogen, because the nitrogen has a lone pair of electrons. This is acid catalysis!
2. At this point, the nitrile carbon is more electropositive than the nitrile nitrogen, so water can act as a nucleophile and bond.
3. To move forward, the nitrogen must be protonated again. So, the water is protonated by abstracting a proton from the intermediate.
4. An intramolecular process occurs to protonate the nitrogen.
5. Since the carbonyl carbon is more electropositive than the carbonyl oxygen, water can attack as a nucleophile.
6. Proton transfer pt 1.
7. Proton transfer pt 2.
8. #"NH"_3# can now leave because its #"pKa"#, which is #36#, is higher than the #"pKa"# of #"H"_2"O"# (#15.7#) if it were to leave. Thus, #bb("NH"_3)# leaves - the equilibrium lies on the side of the weaker acid after the tetrahedral collapse.
9. The protonated carboxylic acid is deprotonated by the ammonia that had left, since it is more basic than water. Water wouldn't deprotonate the carboxylic acid, because hydronium's #"pKa"# is lower than that of ammonium, so the equilibrium lies on the side of ammonia getting protonated.

   So, follow the mechanism, and replace #R# with #"H"#, and you get formic acid (methanoic acid). Ultimately, the first water indicates the first hydrolysis and the second water indicates the second hydrolysis.

   Oxamide is the amide form of oxalic acid. To make it theoretically, stop the acid-catalyzed hydrolysis reaction of cyanogen at the amide. I don't, however think that it's realistically easy.