Question

I need help with percent yield.....


acrylonitrile is produced from the following reaction
2 C3H6 + 2 NH3 + 3 O2 -----> 2 C3H3N + 6 H2O

if 850.0g of C3H6 is mixed with 400.0g NH3 in an unlimited supply of )2, 820.0g of C3H3N was produced. what is the percent yield?


Thank you.

Answer 1

Percent yield = Actual/Theoretical x 100%

Answer 2

this is what I have so far.... I know the formula and started to get the limiting reactant but I'm not sure how .. My high school teacher didn't explain this well.

820g / 15.47 C3H3N = 15.47 ... and 400/ 7.031 NH3


don't know how to finish to get the limiting and I'm not sure how to finish it

Answer 3

Ok first step is to determine what one is the limiting reactant, between 850.0g of C3H6 and 400.0g NH3

Answer 4

To do this we have to convert both to moles,
850.0g C3H6 x 1mole C3H6/12.01(3)+1.01(6)gC3H6 x 2moleC3H3N/2moleC3H6 x 12.01(3)+1.01(3)+14.01gC3H3N/1moleC3H3N holy cow that took forever

Answer 5

usually to find the limiting reactant quick you just convert both to moles of C3H3N and the lower number usually means that the limiting reactant, but to be safe I did the entire problem for 850.0gC3H6

Answer 6

now you do the same for NH3 convert both to moles than moles of C3H3N than grams of C3H3N

Answer 7

make sure to pay attention to the coefficients