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nitrogen gas can be prepared by passing gaseous ammonia over solid copper(ii) oxide at high temperatures. the other products of the reaction are solid acoper and water vapor. if a sample containing 18.1 g of nh3 is reacted with 90.4g of cuo whcih is the limiting reactant? how many grams of n2 will be formed?

Sagot :

10.6g of N2 will be produced as a result when  a sample containing 18.1 g of nh3 is reacted with 90.4g of CuO which is the limiting reactant.

2 NH3(g) + 3 CuO(s) ⟶ N2(g) + 3 Cu(s) + 3 H2O(g)

To calculate moles as

18.1g NH3 X 1mol NH3/17g NH3 = 1.06mol NH3

90.4g CuO x 1mol CuO/79.5g CuO = 1.14mol CuO

Given the limiting reactant 1.06mol NH3 x 3mol CuO/2mol NH3 = 1.59mol CuO is required

Since 1.14mol CuO is present it is limiting

Required is 3mol CuO/2mol NH3 =1.5 and obtained is 1.14mol CuO/1.06mol NH3 = 1.08

Mass of N2 produced = 1.14mol CuO x 1mol N2/3mol CuO x 28g N2/1mol N2= 10.6g N2  

To learn more about limiting reactant click here https://brainly.com/question/14225536

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