1.29g H2 is allowed to react with 10.3g N2 , producing 1.86g NH3.

Question

What is the theoretical yield for this reaction under the given conditions?

What is the percent yield for this reaction under the given conditions?

I have been stucks for days now.

Answer 1

There is a long way and a short way to work these limiting reagent problems. I use the long way.
1. Write and balance the equation.
N2 + 3H2 ==> 2NH3

2a. Convert 1.29 g H2 to moles. moles = grams/molar mass = 1.29/2 = 0.645 moles H2.
2b. Same for N2. moles N2 = 10.3/28 = 0.368.

3a. Using the coefficients in the balanced equation, convert moles H2 to moles of the product, in this case, NH3.
0.645 moles H2 x (2 mol NH3/3 mol H2) = 0.43 x (2/3) = 0.430 moles NH3.
3b. Same for N2.
0.368 N2 x (2 moles NH3/1 mol N2) = 0.368 x (2/1) = 0.736 mole NH3.
3c. You see there are two answers from 3a and 3b; obviously only one can be correct. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. (You should understand why this is true.)
4. Using the smaller value, convert 0.430 moles NH3 to grams. This is the theoretical yield of NH3 in grams. (The 0.430 was the theoretical yield in moles.). 0.430 moles NH3 x (17 g NH3/1 mole NH3) = 7.31 grams NH3 produced. That is the maximum amount that can be produced. What happens to the extra N2. It just sits there. Enough is supplied to react with all of the hydrogen and the excess remains unused.

Then %yield = (actual yield/theoretical yield)*100 = (1.86/7.31)*100 = ?
Print this out. It will work all of your limiting reagent problems.