Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# How many grams are in 3 mol of ##KBr##?

##"360 g"##

Your strategy here will be to use the **molar mass** of potassium bromide, ##"KBr"##, as a conversion factor to help you find the mass of **three moles** of this compound.

So, a compound's **molar mass** essentially tells you the mass of **one mole** of said compound. Now, let's assume that you only have a periodic table to work with here.

Potassium bromide is an ionic compound that is made up of potassium cations, ##"K"^(+)##, and bromide anions, ##"Br"^(-)##. Essentially, one **formula unit** of potassium bromide contains a potassium atom and a bromine atom.

Use to find the **molar masses** of these two . You will find

##"For K: " M_M = "39.0963 g mol"^(-1)##

##"For Br: " M_M = "79.904 g mol"^(-1)##

To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements

##M_"M KBr" = "39.0963 g mol"^(-1) + "79.904 g mol"^(-1) ~~ "119 g mol"^(-)##

So, if **one mole** of potassium bromide has a mas of ##"119 g"##m it follows that **three moles** will have a mass of

##3 color(red)(cancel(color(black)("moles KBr"))) * overbrace("119 g"/(1color(red)(cancel(color(black)("mole KBr")))))^(color(purple)("molar mass of KBr")) = "357 g"##

You should round this off to one , since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs

##"mass of 3 moles of KBr" = color(green)(|bar(ul(color(white)(a/a)"360 g"color(white)(a/a)|)))##