# Question 7f7c0

Mar 12, 2017

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 $\text{16 g}$, 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

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{H") _ 2stackrel(color(blue)(-2))("S") + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")_ 2 -> stackrel(color(blue)(+1)) ("H")_ 2 stackrel(color(blue)(-2))("O") + stackrel(color(blue)(0))("S}}$

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

$\stackrel{\textcolor{b l u e}{+ 4}}{\text{S") ""^(4+) + 4"e"^(-) -> stackrel(color(blue)(0))("S}}$

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 ${\text{64 g mol}}^{- 1}$, it follows that its equivalent mass will be the mass of $\frac{1}{4}$ 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 $\text{16 g}$.