Question #07c92

So, you know that you can get copper metal by roasting a mixture that contains cuprite, #"H"_2"S"#, and copper(II) sulfide, #"CuS"#.

The keys to this problem will be the balanced chemical equations for the reactions that take place.

Both compounds will react with oxygen gas to form copper metal and sulfur dioxide

#"Cu"_2"S"_text((s]) + "O"_text(2(g]) -> color(red)(2)"Cu"_text((s]) + "SO"_text(2(g])#

and

#"CuS"_text((s]) + "O"_text(2(g]) -> "Cu"_text((s]) + "SO"_text(2(g])#

Now, it's worth noting that the question goes out of its way to try and distract you by using purity.

For example, instead of telling that you are heating #"89.0 g"# of mixture, they tell you that you are heating #"100 g"# of mixture that has #11%# impurities.

Likewise, you're interested in how much pure copper is produced by the two reactions, so use the given purity to determine how much copper you have

#75.2color(red)(cancel(color(black)("g impure Cu"))) * "89.4 g Cu"/(100color(red)(cancel(color(black)("g impure Cu")))) = "67.23 g Cu"#

Now, let's assume that the initial sample contains #x# grams of cuprite and #y# grams of copper(II) sulfide. This means that you can write

#x + y = "89.0 g" " " " "color(purple)((1))#

The molar masses of cuprite, copper(II) sulfide, and copper metal, respectively, are

• #"Cu"_2"S " -> " 159.16 g/mol"#
• #"CuS " -> " 95.611 g/mol"#
• #"Cu " -> " 63.546 g"#

If you have #x# grams of cuprite, it follows that you also have

#xcolor(white)(x)color(red)(cancel(color(black)("g"))) * ("1 mole Cu"_2"S")/(159.16color(red)(cancel(color(black)("g")))) = x/159.16"moles Cu"_2"S"#

Likewise, if you have #y# grams of copper(II) sulfide, you will also have

#ycolor(white)(x)color(red)(cancel(color(black)("g"))) * ("1 mole CuS")/(95.611color(red)(cancel(color(black)("g")))) = y/63.546"moles CuS"#

The number of moles of copper will be

#67.23color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "1.058 moles Cu"#

Now focus on the mole ratios that exist between the two species in the mixture and the copper metal.

Notice that you have a #1:color(red)(2)# mole ratio between cuprite and copper metal. This means that each mole of the former will produce #color(red)(2)# moles of the latter.

#x/159.16color(red)(cancel(color(black)("moles Cu"_2"S"))) * (color(red)(2)" moles Cu")/(1color(red)(cancel(color(black)("mole Cu"_2"S")))) = (2x)/159.16"moles Cu"#

Do the same for the #1:1# mole ratio that exists between cuprite and copper metal

#x/95.611color(red)(cancel(color(black)("moles CuS"))) * ("1 mole Cu")/(1color(red)(cancel(color(black)("mole CuS")))) = x/95.611"moles Cu"#

You know that the moles of copper formed by the first reaction and the moles of copper formed by the second reaction must add up to give #"1.058 moles"#.

This means that you have

#(2x)/159.16 + y/95.611 = 1.058" " " "color(purple)((2))#

From this point on, you have a classic system of two equations with two unknowns. Use equation #color(purple)((1))# to write

#x = 89.0- y#

then plug this into equation #color(purple((2))# to get

#(2(89.0-y))/159.16 + y/95.611 = 1.058#

This will be equivalent to

#17019 - 191.22y + 159.16y = 16100.1#

#32.06y = 918.9 implies y = 918.9/32.06 = "28.7 g"#

The value of #x# will be

#x = 89.0 - 28.7 = "60.3 g"#

Therefore, the percent composition of the mixture will be - remember to take the total mass of the mixture, including impurities!

#(28.7color(red)(cancel(color(black)("g"))))/(100.0color(red)(cancel(color(black)("g")))) xx 100 = color(green)("28.7%")# #-># copper(II) sulfide

and

#(60.3color(red)(cancel(color(black)("g"))))/(100.0color(red)(cancel(color(black)("g")))) xx 100 = color(green)("60.3%")# #-># cuprite