# Question #7e89d

##### 1 Answer

#### Explanation:

We're asked to find the **mass, in grams,** of

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 (

Since our goal is to find the number of *grams* of **coefficients of the equation** to find the relative number of **moles** of

#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 **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