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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%