Question

2NH3(g)+3CuO(s)=N2(g)+3Cu(s)+3H2O(g)

if 18.1grams of ammonia reacts with 90.4 grams of copper(II) oxide, which is the limitting reagent?? what is the theorectical yield of nitrogen gasd? if 8.52 g of N2 are formed, what is the % yield of nitrogn gas??

please check if is correct....
A)
18.1gNH3(1 mol NH3/17.03gNH3)(1molN2/2molNH3)= .053 mol N2

90.4gCuO (1molCuO/79.55gCuO) (1molN2/3molCuO)= .379N2, so CuO is the L.R.
B)
0.379molN2 x 28.02gN2/1mol N2= 10.62gN2, ***so, 10.62g N2 is the theoretical yield.
C)
8.52gN2/10.62gN2 x 100= 80.2% yield of N2
please check if I'm doing this problem correct...thankx...

Answer 1

You made a typo for A, it should be 0.053.

For B, I used 28.01 for molar mass N2 and obtained an answer of 10.61 for g N2.
Then 8.52/10.61 *100 = 80.28 which I rounded to 80.3.
Your method is fine. My source for molar mass may be incorrect. I didn't add all of those number myself. Here is the site I used.
http://environmentalchemistry.com/yogi/reference/molar.html#Calculator

Answer 2

a. OK, except the decimal is wrong, typo
b. ok
c. correct.

Answer 3

Thank u!!!!