If 15.0 grams of C3H6, 10.0 grams of oxygen, and 5.00 grams of NH3 are reacted, what mass of acrylonitrile

Question

can be produced assuming 100% yield, what mass of the excess reactants remain?

15g C3H6 (1 mol / 42.08 g) = .35 mol C3H6

.36 mol C3H6 ( 2 mol C3H3N / 2 mol C3H6) = .36 mol C3H3N

10g O2 ( 1 mol/ 32 g O2) = .313 mol O2

.313 mol O2 (2 mol C3H3N/ 3 mol O2) -= .209 mol C3H3N which is the limiting reactant

5 g NH3 ( 1 mol NH3/ 17.03 g) = .294 mol C3H3N

.209 mol C3H3N ( 32 c3H6 / 2 mol C3H3N) = .209 m C3H6

.36mol - .209 mol = .151 mol C3H6

.151 mol C3H6 (42.08 g / 1 mol C3H6 ) = 6.35 g C3H6

Thank you for your help!!

Answer 1

The first thing you need to do is write and balance the equation. then I can help.