Question

2. Write the balanced equation for the formation of ammonia from nitrogen gas and hydrogen gas. Balanced Equation: Given 6 molecules of nitrogen and 12 molecules of hydrogen, make a drawing that represents the reaction container before and after the reaction Before After Now construct a Before-Change-After Table for this reactantur: Bal. Equation: Na - Ha Ha Before: Change: After: Use your representations above to answer the following questions about the reaction a. How many molecules of ammonia can be produced? How do you know? b. Which reactant is in excess? How do you know? c. How many molecules of excess reactant are there? How do you know? Describe what you must look for in a particular reactant mixture to decide which reactant will be in excess (have some left over after the reaction) or which reaction limits the reaction (runs out):

Answer 1

Reaction belween N242 N2(9) +34 2NH21) N2 + 3 42 2NH3 Befere 12 61 change -4 -12 Atlr 8+-8 reactahing wlu 3 mee enl arb One wolecul d N e mdecul N Vapuiren 18 mde culs db Hgurepumd To reaat wit 6 meecule, N, 18 mdacuh & 1 Sna only 12 mdecules 4 tin trept reguired 3 mde cul of H, Produa, 2 mdecul NHR 21 (2 11 (C 8 mdecul nHa Prodye 12 x2 3 G mole cuh d N, takan, dny 4 moe cal seactanted. ene 2mducuu tt unrealted

a) 3 molecules of hydrogen produces 2 molecules of ammonia.

12 molecules produces how many number of ammonia molecules.

Number of ammonia molecules produced = 12 *2/3 = 8 molecules of Ammonia is produced.

b) Nitrogen(N_{​​​​​​2}) is the present in excess. Hydrogen ( H_{​​​​​​2}) is the limiting reagent.

c) 2 moles of nitrogen is present in excess.

d)

First we have to write the balanced equation.

First find how many grams or moles of one Reactant is reactant with how many grams or moles of second reactant.

The number of moles or grams of second reactant required is greater than amount taken, second Reactant is the limiting reagent. If number moles of grams of second reactant required is less than the amount taken, first reactant is the limiting reagent.

In the present problem, I have calculated for 6 moles of nitrogen, how many molecules of Hydrogen is required.

Since Hydrogen required (18 molecules ) is less than that of taken (12 molecules), it is the limiting reagent. Nitrogen is present in excess.