Question #32253
1 Answer
Explanation:
You know that one mole of tennantite contains
#"C"_ color(blue)(12)"As"_4"S"_13#
so the first thing to do here is to figure out how many moles of tennantite you have in that sample.
To do that, use the molar masses of copper, arsenic, and sulfur to find the molar mass of the ore. You will have
#color(blue)(12) color(red)(cancel(color(black)("moles Cu"))) * overbrace("63.546 g"/(1color(red)(cancel(color(black)("mole Cu")))))^(color(purple)("molar mass of Cu")) = "762.552 g " -># from copper
#4color(red)(cancel(color(black)("moles As"))) * overbrace("74.92 g"/(1color(red)(cancel(color(black)("mole As")))))^(color(purple)("molar mass of As")) = "299.68 g " -># from arsenic
#13 color(red)(cancel(color(black)("moles S"))) * overbrace("32.065 g"/(1color(red)(cancel(color(black)("mole S")))))^(color(purple)("molar mass of S")) = "416.845 g " -># from sulfur
The mass of one mole of tennantite, which is essentially what its molar mass is, will thus be
#"762.552 g" + "299.68 g" + "416.845 g" = "1479.08 g"#
Now that you know the mass of one mole of tennantite, use this to find how many moles you have in your sample
#2450 color(red)(cancel(color(black)("g ore"))) * "1 mole ore"/(1479.08color(red)(cancel(color(black)("g ore")))) = "1.6564 moles ore"#
Finally, use the fact that one mole of tennantite contains
#1.6564 color(red)(cancel(color(black)("moles ore"))) * (color(blue)(12)color(white)(a)"moles Cu")/(1color(red)(cancel(color(black)("mole ore")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("19.9 moles Cu")color(white)(a/a)|)))#
The answer is rounded to three sig figs.