If 7.81 g of mg undergoes the following reaction what is the wyield if 6.76 g of magnesium nitride is produced?
3 mg(s) + nale) - mg3n2 (5).

Corrected question:

Since the question has some errors, I copy it with the corrections:

If 7.81 g of Mg undergoes the following reaction what is the % yield, if 6.76 g of magnesium nitride is produced?

Explanation:

1) Mole ratios:

The coefficients on the balanced chemical equation give the mole ratios:

2) Theoretical yield:

       n = mass in grams / molar mass = 7.81 g / 24.305 g/mol = 0.321 mol

       Proportion: 3 mol Mg / 1 mol Mg₃N₂ = 0.321 mol Mg / x

       ⇒ x = 0.321 mol Mg × 1 mol Mg₃N₂ / 3 mol Mg = 0.107 mol Mg₃N₂

3) Percent yield, %

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.