Question

How many molecules of N2 are needed to produce 2.85 g of NF3?

Can someone please explain to me how to do question 3b? Thanks!

Answer 1

Approx. `1xx10^22*"dinitrogen molecules"`........

Explanation:

We interrogate the stoichiometric equation....

`1/2N_2(g) +3/2F_2(g) rarr NF_3(g)`...

Moles of `NF_3-=(2.85*g)/(71.00*g*mol^-1)=0.0401*mol`

And so we need a molar quantity of `(0.0401)/2*mol` with respect to dinitrogen, a mass of `0.562*g`, and a number of `6.022xx10^23*mol^-1xx0.0401*molxx1/2-=1.21xx10^22` with respect to the dinitrogen molecule.... Why did I have to multiply by `1/2`?

Note that our calculation depended on the correct stoichiometric equation, the which informed our ideas of equivalence.....