Question

21.
If 9 moles of O, and 14 moles of N, are placed in a
container and allowed to react according to the
equation :
302 + 2N2 → 2N,03
The reaction proceeds until 3 moles of O, remain,
how many moles of N,O, are present at that instant ?
(a) 6
(b)3
(c) 4
(d) 12​

Answer 1

Answer:

6

Explanation:

using limiting reagents we follow that

given with 9 moles of O2 and having 3 as moles in reaction so 9 ÷3=3

same goes for N2 given with 14 moles but only 2 moles in following reaction so 14÷2=7

as we get O2 as limiting reagent which is used completely in reaction

hence moles of N203 is 6 ( 3 of oxygen × 2 moles of product)