Question #dd10b

1 Answer
Jun 24, 2016

Answer:

Here's what I got.

Explanation:

The first thing to do here is make sure that you're working with a balanced chemical equation, which in this case looks like this

#"Cu"_ 2"S"_ ((l)) + 2"Cu" _ 2"O"_ ((l)) -> 6"Cu"_ ((l)) + "SO"_ (2(g))#

Molten copper(I) sulfide, #"Cu"_2"S"#, will react with molten copper(I) oxide, #"Cu"_2"O"#, to produce molten copper metal and sulfur dioxide, #"SO"_2#.

Now, you know that your reaction must consume #"20 g"# of copper(I) sulfide.

The first thing to do here is use the molar mass of copper(I) sulfide to convert the given mass to moles

#20 color(red)(cancel(color(black)("g"))) * ("1 mole Cu"_2"S")/(159.16color(red)(cancel(color(black)("g")))) = "0.1257 moles Cu"_2"S"#

Use the #1:2# mole ratio that exists between the two reactants to find the number of moles of copper(I) oxide needed to ensure that all the moles of copper(I) sulfide react

#0.1257color(red)(cancel(color(black)("moles Cu"_2"S"))) * ("2 moles Cu"_2"O")/(1color(red)(cancel(color(black)("mole Cu"_2"S")))) = "0.2514 moles CU"_2"O"#

Use the molar mass of copper(I) oxide to convert this to grams

#0.2514 color(red)(cancel(color(black)("moles Cu"_2"O"))) * "143.09 g"/(1color(red)(cancel(color(black)("mole Cu"_2"O")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("36 g Cu"_2"O")color(white)(a/a)|)))#

Do the same for copper metal and sulfur dioxide. The reaction will produce

#0.1257 color(red)(cancel(color(black)("moles Cu"_2"S"))) * "6 moles Cu"/(1color(red)(cancel(color(black)("mole Cu"_2"S")))) = "0.7542 moles Cu"#

This will be equivalent to

#0.7542 color(red)(cancel(color(black)("moles Cu"))) * "63.546 g"/(1color(red)(cancel(color(black)("mole Cu")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("48 g Cu"_2"O")color(white)(a/a)|)))#

I'll leave the last calculation to you as practice.

I'll leave both answers rounded to two sig figs, but keep in mind that you only have one sig fig for the mass of copper(I) sulfide.