Answer to Question #66569 in General Chemistry for Kay

Question

4C3H6 + 6NO --> 4C3H3N + 6H2O + N2

If we have 21.6 g of both of the reactants and the limiting reactant is used up, how much C3H3N will be made? (in grams)

Answer 1

n(C_3 H_6 )=m(C_3 H_6 )/M(C_3 H_6 ) =(21.6 g)/(42.08 g/mol)=0.513 mol
n(NO)=m(NO)/M(NO) =(21.6 g)/(30.01 g/mol)=0.720 mol
n(C_3 H_6 )/4=0.128 mol; (n(NO))/6=0.120
So NO is the limiting reactant.
n(C_3 H_3 N)=4/6 n(NO)=0.480 mol
m(C_3 H_3 N)=n(C_3 H_3 N)×M(C_3 H_3 N)=0.480 mol ×53.06 g/mol=25.5 g
Answer: 25.5g.