Question

Consider the balanced equation for the following reaction:

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)
If 17.3 grams of CuO reacts with 81.4 grams of NH3, how much excess reactant remains?

Answer 1

There will remain 78.9 grams of NH3.

Step-by-step explanation:

Step 1: Data given

Mass CuO = 17.3 grams

Molar mass CuO = 79.545 g/mol

Mass NH3 = 81.4 grams

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

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

Step 3: Calculate moles CuO

Moles CuO = mass CuO/ molar mass CuO

Moles CuO = 17.3 grams / 79.545 g/mol

Moles CuO = 0.217 moles

Step 4: Calculate moles NH3

Moles NH3 = 81.4 grams / 17.03 g/mol

Moles NH3 = 4.78 moles

Step 5: Calculate the limiting reactant

CuO is the limiting reactant. It will completely react. NH3 is in excess. There will react 2/3 * 0.217 = 0.145 moles. There will remain 4.78 - 0.145 = 4.635 moles NH3

Step 6: Calculate mass NH3 remaining

Mass NH3 = 4.635 moles * 17.03 g/mol

Mass NH3 = 78.9 grams

There will remain 78.9 grams of NH3.