At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M and [ NO ] = 0.500 M . N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re‑established?

The answer to the question is the final concentration of NO after equilibrium is re‑established = 0.7143 moles

Explanation:

The reaction is given by

O₂(g) + N₂(g) → 2NO(g)

Kc =

= (0.5^2)/(0.1×0.1) = 25

However, initially, y moles each of O and N gave off x moles each to form 2·x moles of NO, therefore.

Hence we have 2·x = 0.5 or x = 0.25

and y = 0.1+0.25= 0.35

When the concentration is increased to 0.8, we have each mole of NO giving out 2·a moles while each O₂ and N₂  increases by a moles which gives

(0.8-2·a)^2/(0.1+a)^2 = 25

Taking square roots of both sides and cross multiplying, we have 7×a=0.3

Which gives a = 0.04286 moles.

Therefore the number of moles of NO at equilibrium

= 0.8-2×0.04286 = 0.7143 moles

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g