Question

If 14.5 moles of Cu and 44.8 moles of HNO3 are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? 3 Cu + 8 HNO3 ? 3Cu(NO3)2 + 2NO + 4H2O

Answer 1

LEt's write again the overall balanced reaction:

3Cu + 8HNO₃ -----------> 3Cu(NO₃)₂ + 2NO + H₂O

we have 14.5 moles of Cu reacting with 44.8 moles of HNO₃. So, let's find out who's the limitant reactant:

3 moles Cu --------> 8 moles HNO₃

14.5 moles Cu ---------> X

X = 8 moles HNO₃ * (14.5 moles Cu / 3 moles Cu) = 38.66 moles HNO₃

We have 44.8 moles HNO₃ so the limitant reactant is Cu and the excess is HNO₃.

When the reaction is complete, the 14.5 moles of Cu react with only38.66 moles of HNO₃. The remaining moles of HNO₃ are:

moles of HNO₃ remaining = 44.8 - 38.66 = 6.14 moles of HNO₃ remaining.

Hope this helps