Question #6584b

1 Answer
Sep 15, 2017

Answer:

#"4 moles KClO"_3#

Explanation:

The key here is the balanced chemical equation that describes this decomposition reaction

#2"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(-> ) 2"KCl"_ ((s)) + 3"O"_ (2(g))#

Take a look at the coefficients added to the chemical species that take part in the reaction.

In this case, you know that for every #2# moles of potassium chlorate that undergo decomposition, you get #2# moles of potassium chloride and #3# moles of oxygen gas.

  • #2"KClO"_3 -> "2 moles of KClO"_3#
  • #"2KCl " -> " 2 moles KCl"#
  • #3"O"_2 -> "3 moles O"_2#

This means that in order for the reaction to produce #6# moles of oxygen gas, you need to have

#6 color(red)(cancel(color(black)("moles O"_2))) * overbrace( "2 moles KClO"_3/(3color(red)(cancel(color(black)("moles O"_2)))))^(color(blue)("given by the balanced chemical equation")) = color(darkgreen)(ul(color(black)("4 moles KClO"_3)))#