When 14.0 grams of nitrogen is combined with excess hyrdogen in a lab, 16.1 grams of ammonia (nh3) forms. write and balance the reaction.

The reaction between Nitrogen and Hydrogen to form ammonia gas is given by the equation;

N₂ + 3H₂ → 2NH₃

From the equation 1 mole of nitrogen gas (28 g) requires 3 moles of hydrogen gas (6 g) to form 2 moles of Ammonia gas (34 g).

Therefore;

Moles of Nitrogen = 14/28 = 0.5 mole

Moles of Ammonia = 16.1 /17 = 0.947 ≈ 1 moles

Therefore;

0.5 mole of nitrogen gas (14 g) requires 1.5 moles of hydrogen gas (3 g) to form 1 mole of Ammonia gas (17 g).

The balanced equation would be;

N₂ + 3H₂ → 2NH₃