Question #a5001
1 Answer
Explanation:
Take a look at the balanced chemical equation for this reaction
#"CS"_text(2(l]) + color(red)(2)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"SO"_text(2(g])#
Notice the
The problem tells you that you need to determine how many grams of oxygen would be needed to react with
Right from the start, you know that the reaction must consume
#7.674 color(red)(cancel(color(black)("moles CS"_2))) * (color(red)(3)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole CS"_2)))) = "21.022 moles O"_2#
To find out how many grams of oxygen would contain this many moles, use oxygen gas' molar mass
#21.022 color(red)(cancel(color(black)("moles O"_2))) * overbrace("32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(purple)("molar mass of O"_2)) = color(green)(|bar(ul(color(white)(a/a)"736.7 g O"_2color(white)(a/a)|)))#