PLEASE HELP ME One method to produce nitrogen in the lab is to react ammonia with copper (II) oxide: NH3(g) + CuO(s) ---> Cu(s) + H2O(l) + N2(g). After using 40.0 grams of NH3, 15.5 grams of N2 are produced. What is the percent yield of nitrogen in the reaction?

47.05% is the percent yield of nitrogen in the reaction.

Explanation:

heoretical yield of nitrogen gas = x

Moles of ammonia =  

According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.

Then 2.3529 mol of ammonia will give:

of nitrogen gas

Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g

Experiential yield of nitrogen gas = 15.5 g

Percentage yield:

hope that help

47.05% is the percent yield of nitrogen in the reaction.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.