Question

Question #9ca2c

Answer 1

`1.0 * 10^2"g Cu"_2"S"`

Explanation:

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

`color(red)(16)"Cu"_ ((s)) + "S"_ (8(s)) -> color(blue)(8)"Cu"_ 2"S"_((s))`

Notice that the reaction consumes `color(red)(16)` moles of copper metal and produces `color(blue)(8)` moles of copper(I) sulfide for every mole of octasulfur, `"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, `"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 `color(red)(16) : color(blue)(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)|)))`