Question #7e89d

We're asked to find the mass, in grams, of #"Mg"_3"N"_2# that form when #2.0# #"mol N"_2# and #8.0# #"mol Mg"# react.

We're given the chemical equation:

#ul(3"Mg"(s) + "N"_2(g) rarr "Mg"_3"N"_2(s)#

Since we're given the quantities of more than one reactant, we have to find the limiting reactant.

To do this, we take the number of moles of each reactant, and divide it by the respective coefficient of that reactant. Whichever reactant has the lower number is limiting.

We're given:

#2.0# #"mol N"_2#

#8.0# #"mol Mg"#

Now we divide by their coefficients:

#"N"_2:# #(2.0color(white)(l)"mol")/(1color(white)(l)"(coefficient)") = ul(2.00)#

#"Mg":# #(8.0color(white)(l)"mol")/(3color(white)(l)"(coefficient)") = ul(2.67#

Since nitrogen's number is lower, #"N"_2# is the limiting reactant, so we will use nitrogen's mole value (#2.0# #"mol"#) during our calculations.

Since our goal is to find the number of grams of #"Mg"_3"N"_2#, let's use the coefficients of the equation to find the relative number of moles of #"Mg"_3"N"_2# that can form:

#2.0cancel("mol N"_2)((1color(white)(l)"mol Mg"_3"N"_2)/(1cancel("mol N"_2))) = color(red)(ul(2.0color(white)(l)"mol Mg"_3"N"_2#

Now, we use the molar mass of #"Mg"_3"N"_2# (#100.928# #"g/mol"#) to find the number of grams:

#color(red)(2.0)cancel(color(red)("mol Mg"_3"N"_2))((100.928color(white)(l)"g Mg"_3"N"_2)/(1cancel("mol Mg"_3"N"_2))) = color(blue)(ulbar(|stackrel(" ")(" "2.0xx10^2color(white)(l)"g Mg"_3"N"_2" ")|)#

rounded to #2# significant figures.