Here’s what I got.

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 , but keep in mind that you only have one sig fig for the mass of copper(I) sulfide.