17.85 g of Mg undergoes the following reaction, what is the %s yield if 6.76 g of magnesium nitride is produced? 3 Mg(s)+N2(g)—->Mg 3 N 2 (s)

Answer 1

Before we start, we want to make sure that the equation is balanced, and it looks like the question writer has already balanced this equation for us.

We are given the actual yield of the equation as 6.76 grams of magnesium nitride.

The equation to determine percent yield is as follows:

% yield = (Actual yield / Theoretical yield) x 100%

To determine theoretical yield, let us start out with the amount of magnesium reactant that will undergo the reaction. Using stoichiometry and our mole coefficients, we can determine the following.

17.85 grams of Mg x 1 mol Mg x 1 mol Mg₃N₂ x 100 grams Mg₃N_{2 =} 24.480 grams of Mg₃N₂

24.305 grams 3 mol Mg 1 mol Mg₃N₂

To determine our percent yield:

% yield = (6.76 grams Mg₃N₂ / 24.480 grams Mg₃N₂) x 100% = 27.61% yield