Question

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?

Answer 1

copper(ll)oxide

Step-by-step 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