Nitrogen gas can be prepared by passing ammonia over solid copper (ii) oxide according to the equation2 nh3 + 3 cuo (s) --> n2 (g) + 3 cu (s) + 3 h2o (g)suppose 18.1 grams of ammonia is reacted with 90.4 grams cuo.

a) what is the theoretical yield of nitrogen in grams? b) what is the theoretical yield of copper in grams? c) which reactant is limiting?

The theoretical yield of nitrogen is 10.64 grams

The theoretical yield of copper is 72.44 grams

CuO is the limiting reactant.

Explanation:

Step 1: Data given

Mass of NH3 = 18.1 grams

Mass of CuO = 90.4 grams

Molar mass NH3 = 17.03 g/mol

Molar mass CuO = 79.545 g/mol

Molar mss of Cu = 63.546 g/mol

Step 2: The balanced equation

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

Step 3: Calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 18.1 grams / 17.03 g/mol

Moles NH3 = 1.06 moles

Step 4: Calculate moles CuO

Moles CuO = 90.4 grams / 79.545 g/mol

Moles CuO = 1.14 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 we need 3 moles CuO to produce 1 mol N2, 3 moles Cu and 3 moles H2O

CuO is the limiting reactant. It will completely be consumed (1.14 moles).

NH3 is in excess. There will react 2/3 * 1.14 = 0.76 moles

There will remain 1.06 -0,76 = 0.30 moles of NH3

Step 6: Calculate moles of N2 and Cu

For 2 moles NH3 we need 3 moles CuO to produce 1 mol N2, 3 moles Cu and 3 moles H2O

For 1.14 moles of CuO we'll have 1.14/ 3 = 0.38 moles of N2 and 1.14 moles of Cu

Step 7: Calculate mass of products

Mass N2 = moles N2 * molar mass N2

Mass N2 = 0.38 moles * 28 g/mol

Mass N2 = 10.64 grams

Mass  Cu = 1.14 moles * 63.546 g/mol

Mass Cu = 72.44 grams grams

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.