# How much potassium bromide, "KBr", is produced if 20.0g of potassium bromate, "KBrO"_3, decomposes?

Jun 27, 2016

$\text{14.3 g KBr}$

#### Explanation:

Your starting point here will be the balanced chemical equation that describes this decomposition reaction.

Potassium bromate, ${\text{KBrO}}_{3}$, will undergo thermal decomposition to form potassium bromide, $\text{KBr}$, and oxygen gas, ${\text{O}}_{2}$.

$2 {\text{KBrO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KBr"_ ((s)) + 3"O}}_{2 \left(g\right)}$

Notice that it takes $2$ moles of potassium bromate to form $2$ moles of potassium bromide. This tells you that the reaction will produce as many moles of potassium bromide as you have moles of potassium bromate that undergo decomposition.

Simply put, the two compounds are in a $1 : 1$ mole ratio.

You can convert this mole ratio to a gram ratio by using the molar masses of the two compounds

M_("M KBrO"_3) = "167.0 g mol"^(-1)

M_("M KBr") = "119.0 g mol"^(-1)

This means that a $1 : 1$ mole ratio is equivalent to a $167.0 : 119.0$ gram ratio.

All you have to do now is use this gram ratio to determine how many grams of potassium bromide are produced when $\text{20.0 g}$ of potassium bromate undergo decomposition

20.0 color(red)(cancel(color(black)("g KBrO"_3))) * "119.0 g KBr"/(167.0color(red)(cancel(color(black)("g KBrO"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("14.3 g KBr")color(white)(a/a)|)))

The answer is rounded to three sig figs.