Question #7f7c0

1 Answer
Mar 12, 2017

Answer:

Here's why that is the case.

Explanation:

The first thing to note here is that equivalent mass is not unitless. In this case, the equivalent mass of sulfur dioxide is actually #"16 g"#, not #16#.

Now, the trick here is to realize that you're dealing with a redox reaction, so start by assigning oxidation numbers to the atoms that take part in the reaction

#stackrel(color(blue)(+1))("H") _ 2stackrel(color(blue)(-2))("S") + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")_ 2 -> stackrel(color(blue)(+1)) ("H")_ 2 stackrel(color(blue)(-2))("O") + stackrel(color(blue)(0))("S")#

Notice that sulfur is being oxidized and reduced.

The oxidation half-reaction looks like this

#stackrel(color(blue)(-2))("S") ""^(2-) -> stackrel(color(blue)(0))("S") + 2"e"^(-)#

The reduction half-reaction looks like this

#stackrel(color(blue)(+4))("S") ""^(4+) + 4"e"^(-) -> stackrel(color(blue)(0))("S")#

Now, notice that when hydrogen sulfide gets oxidized, the sulfur atom loses #2# electrons. On the other hand, when sulfur dioxide gets reduced, the sulfur atom gains #4# electrons.

In other words, when this reaction takes place, every mole of hydrogen sulfide will lose #2# moles of electrons and every mole of sulfur dioxide will gain #4# moles of electrons.

Now, in the context of a redox reaction, the equivalent mass of a compound is taken to be the mass of said compound that can either supply or react with #1# mole of electrons.

Since every sulfur dioxide reacts with #4# moles of electrons, you can say that #1# mole of electrons will need

#1 color(red)(cancel(color(black)("mole e"^(-)))) * "1 mole SO"_2/(4color(red)(cancel(color(black)("moles e"^(-))))) = 1/4color(white)(.)"moles SO"_2#

Since sulfur dioxide has a molar mass of about #"64 g mol"^(-1)#, it follows that its equivalent mass will be the mass of #1/4# moles of sulfur dioxide

#1/4 color(red)(cancel(color(black)("moles SO"_2))) * "64 g"/(1color(red)(cancel(color(black)("mole SO"_2)))) = "16 g"#

And there you have it -- the equivalent mass of sulfur dioxide for this reaction is #"16 g"#.