# Question 218f2

Jun 12, 2016

$\text{32 g}$

#### Explanation:

A substance's equivalent mass actually depends on the reaction in which said substance is involved.

See this Wikipedia article for more details on how equivalent mass works.

In your case, the equivalent mass of sulfur, $\text{S}$, will be the mass that combines with $\text{35.5 g}$ of chlorine, $\text{Cl}$, which is the mass of one mole of chlorine.

For every mole of sulfur dichloride, ${\text{SCl}}_{2}$, you get

• one mole of sulfur, $1 \times \text{S}$
• two moles of chlorine, $2 \times \text{Cl}$

Since sulfur has a molar mass of approximately ${\text{32 g mol}}^{- 1}$, you can say that a mole of sulfur dichloride contains

• $1 \times \text{32 g}$ of sulfur
• $2 \times \text{35.5 g" = "71 g}$ of chlorine

So, if $\text{32 g}$ of sulfur combined with $\text{71 g}$ of chlorine, it follows that $\text{35.5 g}$ of chlorine combined with

35.5 color(red)(cancel(color(black)("g Cl"))) * "32 g S"/(71color(red)(cancel(color(black)("g Cl")))) = "16 g S"

This is why the equivalent mass of sulfur in sulfur dichloride is said to be $\text{16 g}$.

Now, in disulfur dichloride, ${\text{S"_2"Cl}}_{2}$, one mole contains

• two moles of sulfur, $2 \times \text{S}$
• two moles of chlorine ,$2 \times \text{Cl}$

This time, you have $2 \times \text{32 g}$ of sulfur combine with $2 \times \text{35.5 g}$ of chlorine, which means that $\text{35.5 g}$ of chlorine will combine with

35.5color(red)(cancel(color(black)("g Cl"))) * (2 xx "32 g S")/(71color(red)(cancel(color(black)("g Cl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("32 g S")color(white)(a/a)|)))#

Therefore, the equivalent mass of sulfur in disulfur dichloride will be $\text{32 g}$.

You'll sometimes see this referred to as the mass of one equivalent of sulfur, which in this context is simply another term used to denote the mass of sulfur that combines with $\text{35.5 g}$ of chlorine.