# Question 9ca2c

May 29, 2016

$1.0 \cdot {10}^{2} \text{g Cu"_2"S}$

#### Explanation:

Start by writing out the balanced chemical equation that describes this synthesis reaction

$\textcolor{red}{16} {\text{Cu"_ ((s)) + "S"_ (8(s)) -> color(blue)(8)"Cu"_ 2"S}}_{\left(s\right)}$

Notice that the reaction consumes $\textcolor{red}{16}$ moles of copper metal and produces $\textcolor{b l u e}{8}$ moles of copper(I) sulfide for every mole of octasulfur, ${\text{S}}_{8}$, that takes part in the reaction.

The first thing to do here is use the molar masses of the two reactants to determine how many moles of each you're mixing

82 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "1.29 moles Cu"

25 color(red)(cancel(color(black)("g"))) * "1 mole S"_8/(256.5color(red)(cancel(color(black)("g")))) = "0.0975 moles S"_8

Notice that the molar mass of octasulfur is eight times that of elemental sulfur, $\text{S}$.

So, do you have enough moles of copper metal to ensure that all the moles of octasulfur take part in the reaction?

That many moles of octasulfur would require

0.0975color(red)(cancel(color(black)("moles S"_8))) * (color(red)(16)color(white)(a)"moles Cu")/(1color(red)(cancel(color(black)("mole S"_8)))) = "1.56 moles Cu"

Since you only have $1.29$ moles of copper metal available, it follows that copper will act as a limiting reagent.

That means that copper will be completely consumed by the reaction before all the moles of octasulfur get the chance to react.

Now, notice that you have a $\textcolor{red}{16} : \textcolor{b l u e}{8}$ mole ratio between copper and copper(I) sulfide. Since the reaction will consume $1.29$ moles of copper, it follows that it will also produce

1.29 color(red)(cancel(color(black)("moles Cu"))) * (color(blue)(8)color(white)(a)"moles Cu"_2"S")/(color(red)(16)color(red)(cancel(color(black)("moles Cu")))) = "0.645 moles Cu"_2"S"

To determine how many grams of copper(I) sulfide would contain this many moles, use the compound's molar mass

0.645 color(red)(cancel(color(black)("moles Cu"_2"S"))) * "159.16 g"/(1color(red)(cancel(color(black)("mole Cu"_2"S")))) = "102.7 g"

Rounded to two sig figs, the answer will be

"mass of Cu"_2"S" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.0 * 10^2"g")color(white)(a/a)|)))#