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

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

1 Answer
Dec 16, 2017

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.....