Question #d9a4f

1 Answer
Aug 6, 2016

Answer:

Here's what I got.

Explanation:

Your starting point here is the balanced chemical equation that describes this decomposion reacton

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

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

To convert these mole ratios to grams ratios, use the molar masses of three chemical species that take part in your reaction.

#M_("M KClO"_3) = "122.55 g mol"^(-1)#

#M_("M KCl") = "74.55 g mol"^(-1)#

#M_("M O"_2) = "32.0 g mol"^(-1)#

You can thus say that every

#2 color(red)(cancel(color(black)("moles KClO"_3))) * "122.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))) = "241.5 g"#

of potassium chlorate that undergo decomposition will produce

#2 color(red)(cancel(color(black)("moles KCl"))) * "74.55 g"/(1color(red)(cancel(color(black)("mole KClO"_3)))) = "149.1 g"#

of potassium chloride and

#3 color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = "96.0 g"#

of oxygen gas. Use these gram ratios to calculate how many grams of potassium chloride and oxygen gas will result from the decomposition of #"49 g"# of potassium chlorate

#49 color(red)(cancel(color(black)("g KClO"_3))) * "149.1 g KCl"/(241.5 color(red)(cancel(color(black)("g KClO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("30. g KCl")color(white)(a/a)|)))#

#49 color(red)(cancel(color(black)("g KClO"_3))) * "96.0 g O"_2/(241.5 color(red)(cancel(color(black)("g KClO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("19 g O"_2)color(white)(a/a)|)))#

Both answers are rounded to two sig figs, the number of sig figs you have for the mass of potassium chlorate.