Question #8c336

Conversion factors are always set up like this

#"what you need"/"what you have"#

In this case, you have the balanced chemical equation

#"Mg"_ 3"N"_ (2(s)) + color(blue)(6)"H"_ 2"O"_ ((l)) -> 3"Mg"("OH")_ (2(s)) + color(purple)(2)"NH"_ (3(g))#

Notice that you have a #color(blue)(6) : color(purple)(2)# mole ratio between water, a reactant, and ammonia, one of the products.

You must convert the number of moles of water that took place in the reaction to the number of moles of ammonia produced by the reaction, so the conversion factor will look like this

#(color(purple)(2)color(white)(.)"moles NH"_3)/(color(blue)(6)color(white)(.)"moles H"_2"O") color(white)( aacolor(black)( larr " what you need")/(color(black)(larr" what you have"))#

You can simplify this ratio to

#(color(purple)(2)color(white)(.)"moles NH"_3)/(color(blue)(6)color(white)(.)"moles H"_2"O") = (color(red)(cancel(color(black)(2))) * "1 mole NH"_3)/(color(red)(cancel(color(black)(2))) * "3 moles H"_2"O") = "1 mole NH"_3/("3 moles H"_2"O")#

Let's say that the reaction consumes #6# moles of water. You can use the mole ratio that exists between the two compounds as a conversion factor to say that the reaction will produce

#6 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole NH"_3/(3color(red)(cancel(color(black)("moles H"_2"O")))) = "2 moles NH"_3#

So remember, as far as conversion factors go, what you need is always added to the top, i.e. in the numerator, and what you have is always added to the bottom, i.e. in the denominator.