3.0 g of hydrogen + 24 g of oxygen = 27 g of waterThis leaves 12 g − 3.0 g = 9.0 g of hydrogen uncombined.Typical Problem (by Volume)
A mixture of 8 milliliters of hydrogen and 200 milliliters of air is placed in a
steam-jacketed eudiometer, and a spark is passed through the mixture. What will
be the total volume of gases in the eudiometer?
Because this is a combination by volume, 8 mL of hydrogen require 4 mL of
oxygen. (Ratio H 2 : O 2 by volume = 2 : 1.)
The 200 mL of air is approximately 21% oxygen. This will more than supply
the needed oxygen and leave 196 mL of the air uncombined.
The 8 mL of hydrogen and 4 mL of oxygen will form 8 mL of steam since the
eudiometer is steam-jacketed and keeps the water formed in the gaseous state.
(Ratio by volume of hydrogen : oxygen : steam = 2 : 1 : 2)
TOTAL VOLUME = 196 mL of air + 8 mL of steam = 204 mLHEAVY WATER
A small portion of water is called “heavy” water because it contains an isotope of
hydrogen, deuterium (symbol D), rather than ordinary hydrogen nuclei. Deuterium
has a nucleus of one proton and one neutron rather than just one proton. Another
isotope of hydrogen is tritium. Its nucleus is composed of two neutrons and one
proton. Both of these isotopes have had use in the nuclear energy field.
HYDROGEN PEROXIDE
The prefix per- indicates that this compound contains more than the usual oxide.
Its formula is H 2 O 2. It is a well-known bleaching and oxidizing agent. Its
electron-dot formula is shown in Figure 28.