When gaseous ammonia is passed over solid copper (Il) oxide at high temperatures, nitrogen gas is formed. 2NH3 (g) + 3CuO (s) - --> N2 (g) + 3Cu (s) + 3H20 (g) What is the limiting reagent when 34 grams of ammonia form 26 grams of nitrogen in a reaction that runs to completion?

copper (ll) oxide

Explanation:

Limit reagent is a reagent that finished thereby stopping the reaction.

2NH3 (g) + 3CuO (s) - --> N2 (g) + 3Cu (s) + 3H20 (g)

molar mass of of ammonia = 17.031 * 2 (stiochiometry mole) = 34.06 g/mol

molar mass of CuO = 79.545 * 3 = 238.635 g/mol

molar mass of Nitrogen gas = 28.0134 g/mol

34.06 g of NH3 need 238.635 of CuO to produce 28.0134 of Nitrogen gas

238.635g produce 28.0134 g

x gram of CuO produce 26 gram of Nitrogen gas

x gram = (26 * 238.635) / 28.0134 = 221.484 g

also

34.06 g of ammonia produces 28.0134g of nitogen gas

y gram of ammonia produce 26g

y gram = (26 * 34.06) / 28.0134 g = 31.61 g

but the 34 g of ammonia was used in the reaction and 31.61 reacted leaving and 2.39 g of ammonia gas

The limiting reagent therefore is copper (ll) oxide