Question #d5e50

The balanced chemical equation that describes this synthesis reaction looks like this

#color(red)(3)"Mg"_ ((s)) + "N"_ (2(g)) -> "Mg"_ 3"N"_(2(s))#

Notice that in order to produce #1# mole of magnesium nitride, the reaction must consume #color(red)(3)# moles of magnesium metal and #1# mole of nitrogen gas.

You already know how many moles of magnesium metal you have at your disposal, so use the #color(red)(3):1# mole ratio that exists between the two reactants to determine how many moles of nitrogen gas are needed

#0.45color(red)(cancel(color(black)("moles Mg"))) * "1 mole N"_2/(color(red)(3)color(red)(cancel(color(black)("moles Mg")))) = "0.15 moles N"_2#

All you have to do now is use the molar mass of nitrogen gas to determine how many grams of nitrogen gas would contain that many moles

#0.15 color(red)(cancel(color(black)("moles N"_2))) * "28.01 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(|bar(ul(color(white)(a/a)"4.2 g N"_2color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of magnesium.