Question

what mass of nitrogen will be required to produce 6.022×10^24 molecule of ammonia?​

Answer 1

Answer:

You got what…10 moles of ammonia required? One mole of stuff specifies 6.022×10236.022×1023 individual items of stuff… And of course we know (do we?) that 6.022×10236.022×1023 12C12C atoms have a mass of 12.0⋅g12.0·g precisely.

Now here we got a molar quantity WITH RESPECT to AMMONIA of…

6.022×10246.022×1023⋅mol−1=10⋅mol6.022×10246.022×1023·mol−1=10·mol .

And so given the stoichiometric equation…

12N2+32H2(g)→NH3(g)12N2+32H2(g)→NH3(g)

And so we need a 5 mole quantity of dinitrogen….i.e. approx. 70⋅g70·g .

Note that this is one of the most important industrial reactions. Without dinitrogen fixation there is no nitrogenous fertilizer, and NO food…

Good question by the way…you don’t really need a calculator, and yet it stills test the ideas of stoichiometry and equivalence.