When gaseous ammonia is passed over solid copper(il) oxide at high temperatures, nitrogen gas is formed. 2nh3(g) + 3cuo(> 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

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