Consider the following balanced equation:
3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)
If 70.8 moles of CuO(s) and 40.1 moles of NH3(g) are allowed to react, what is the theoretical yield of Cu(s) in moles?

The theoretical yield of Cu(s) in moles is 60.15 moles

Explanation:

Step 1: Data given

Number of moles CuO = 70.8 moles

Number of moles NH3 = 40.1 moles

Molar mass CuO = 79.545 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

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

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

NH3 is the limiting reactant. It will completely be consumed (40.1 moles). CuO is in excess. There will react 3/2 * 40.1 = 60.15 moles

There will remain 70.8 - 60.15 = 10.65 moles CuO

Step 3: Calculate moles Cu

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

For 40.1 moles NH3 we'll have 60.15 moles Cu

The theoretical yield of Cu(s) in moles is 60.15 moles

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm