Question

By using 21 gram N2 and 8 gram H2 how much amount of ammonia we prepared.​

Answer 1

Explanation:

How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2? So, 68 g NH3 can be produced because N2 is the limiting reactant and you will run out of it first.

Answer 2

Answer:N2(g) + 3H2-> 2NH3(g) This is the balanced equation

Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced

grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

Explanation: