Question

In the synthesis of ammonia, if `4.88xx10^-4*g` dihydrogen react, how many individual ammonia molecules are formed?

Answer 1

`1.55xx10^-4*molxx6.022xx10^23*mol^-1*"ammonia molecules"`

Explanation:

We need a stoichiometric equation to represent the formation of ammonia:

`1/2N_2(g) + 3/2H_2(g) stackrel"catalysis"rarrNH_3(g)`

We have a molar quantity with respect to dihydrogen of...

`(4.88xx10^-4*g)/(2.01*g*mol^-1)=2.32xx10^-4*mol`

And the given stoichiometry requires that we form 2/3 of an equivalent with respect to ammonia, i.e. `2.32xx10^-4*molxx2/3*"equiv"=1.55xx10^-4*mol`

Now we know that one mole of stuff specifies `6.022xx10^23` individual molecules....

And so (finally) we take the product....

`1.55xx10^-4*molxx6.022xx10^23*mol^-1*"ammonia molecules"`

`=??`