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Please solve this!! I won’t be able to talk since I’m in school but just end it if im not there and I’ll give you a 5 stars when I come back to look at it.

Please solve this I wont be able to talk since Im in school but just end it if im not there and Ill give you a 5 stars when I come back to look at it class=


Answer :

[tex]Mg_3N_2+K_2O\Rightarrow\text{MgO}+K_3N[/tex]

Step 1) We must balance our equation.

[tex]Mg_3N_2+3K_2O\Rightarrow3\text{MgO}+2K_3N[/tex]

(we will work with this equation)

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Step 2) The limiting reactant = K2O

We need to find the mass of K3N, so we need its molar mass (use the periodic table):

Molar Mass K3N = 130 g/mol (approx.)

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Step 3) Use the stoichiometry

3 x 1 mole of K2O ----------- 2 x 130 g of K3N

14 moles of K2O----------- X

[tex]X\text{ = }\frac{14\text{ moles x 2x130 g}}{3\text{ x 1mole}}=\text{ 1213.13 g =1200 g (approx.) }[/tex]

Answer: Mass of K3N = 1200 g (approx.)

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