what mass of nitrogen will be required to produce 6.022 * 10^24 molecules of ammonia by following reaction? N2+3H2------>2NH3
Answers
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.