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

17.85gMg / 24.31g=0.734 mol Mg

3 mol Mg forms 1 mol Mg3N2

so, 0.734molMg /3 = 0.245 mol Mg3N2

molar mas Mg3N2

3(Mg)+2(N)

3(24.31)+2(14.01)=100.95g/1mol

0.245mol(100.95g/mol)

=24.73 g Mg3N2 theoretically forms

6.76 g actually forms

%yield=(actual/theoretical)100

=(6.76/24.73)100

=27.33% yield