How many grams of ammonia (NH3) are produced when you react 4.76 grams of hydrogen (H2)

Answer: hre

Explanation:

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.

Mass of NH3 = ?

Mass of H2 = 4.76g

Molar mass of H2

=1.00784g/mol*2 = 2.0158

no of moles of H2 = 4.76 / 2.0158

No of mole of H2 = 2.3613mol

Equation of the reaction =

N2 (g) + 3H2 (g) = 2NH3 (g)

3 mole of H2 gas react to form 2 moles of NH3 (g)

Then 2.3613mol of H2 (g) will produce 2.3613*2/3 = 1.5742mol of NH3 (g)

Molar mass 17.031g/mol

Mass of NH3 produced = Mass * Molar mass = 1.5742*17.031

= 26.810g

Mass of NH3 produced is 26.810g