In reaction NH3 →3H2 + N2, how many moles of N2 formed if 2.81 g NH3 dissociate? Show work!

0.0826 mol (corrected to 3 sig. fig.)

Explanation:

First, balance the equation:

2NH3 →3H2 + N2

Take the atomic no. of N=14.0, and H=1.0,

no. of moles = mass / molar mass

So, no. of moles of NH3 dissociated = 2.81 / (14.0+1.0x3)

= 0.165294117mol

From the equation, the mole ratio of NH3:N2 = 2:1, meaning for every 2 moles of NH3 dissociated, one mole of N2 is formed.

So, using this ratio, the no. of moles of N2 formed will be 0.165294117 / 2

=0.0826 mol (corrected to 3 sig. fig.)

what is the question?

pv=nrt

explanation: