Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
n2(g) + 3h2(g) ? 2nh3(g)
1. what is the maximum mass of ammonia that can be produced from a mixture of 163.3 g of n2 and 38.77 g of h2? g
2. which element would be completely consumed? (enter nitrogen or hydrogen)
3. what mass of the starting material would remain unreacted?
look at how many grams of ammonia the n2 would produce and how many grams of ammonia the h2 would produce.
watch the mole ratios!
whichever produces the smallest amount of ammonia is the limiting reactant.
hint: after you have found which starting material was used up completely, you know that some of the
other starting material is still unreacted. subtract the reacted material (which you must calculate) from the
original amount you had in the beginning, thus finding how much remains unreacted

 1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

pla·gia·rism

/ˈplājəˌrizəm/

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noun

the practice of taking someone else's work or ideas and passing them off as one's own.

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