Answer:
There will remain 78.9 grams of NH3.
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.
