The following reaction is investigated: 2N2O(g) N2H4(g) <---> 3N2(g) 2H2O(g) Initially there are 0.100 mol of N2O and 0.25 mol of N2H4, in a 10.0-L container. If there are 0.059 mol of N2O at equilibrium, how many moles of N2 are present at equilibrium

6.2 × 10^-2

Explanation:

Given the reaction:

2N2O(g) N2H4(g) <---> 3N2(g) 2H2O(g)

Initial: N2O = 0.1, N2H4 = 0.25-- 0 - - - - - 0

Then: N20 = -2x, N2H4= - x - - - +3x, +2x

Equ : 0.1 - 2x ; 0.25 - x ; +3x ; +2x

At equilibrium :

Add both:

N2O = 0.1 - 2x ;

N2H4 = 0.25 - x;

3N2 = 3x

2H2O = 2x

Moles of N2O at equilibrium = 0.059

Then;

0.1 - 2x = 0.059

-2x = 0.059 - 0.1

-2x = - 0.041

x = 0.041 / 2

x = 0.0205

Moles of N2 present at equilibrium ;

3N2 = 3x

3N2 = 3(0.0205)

= 0.0615

= 0.062 = 6.2 × 10^-2

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