Question

If you have 1.00 mole of each reactant, which will be the limiting reactant?

Answer 1

The reactants are the one to the left side, so they are Cu and HNO₃.

Their coefficients are 3 and 8, so for each 3 Cu that reacts, we will need 8 HNO₃

The Limiting reactant is the one that we have less considering the proportions they will react.

Let's see the two possible cases.

If we have 1 mol of Cu, we need to divide it by its coefficient and multiply it by the coefficient of HNO₃:

HNO₃ --- Cu

8 --- 3

`\begin{gathered} (n_(HNO_3))/(8)=(n_(Cu))/(3) \\ n_(HNO_3)=(8n_(Cu))/(3)=(8\cdot1)/(3)mol=(8)/(3)mol=2.666\ldots mol \end{gathered}`

This means that we need about 2.7 mol of HNO₃ to react all the 1 mol of Cu we have.

Since we have 1 mol of HNO₃, we don't have enough.

If we make the other way around, we will get the following:

1 mol of HNO₃, so we need:

Cu --- HNO₃

3 --- 8

`\begin{gathered} (n_(Cu))/(3)=(n_(HNO_3))/(8) \\ n_(Cu)=\frac{3n_{HNO_(3)}}{8}=(3\cdot1)/(8)mol=0.375mol \end{gathered}`

So, we would need 0.375 mol of Cu to react with all 1 mol of HNO₃ we have, and since we have 1 mol of Cu, this is enough.

So, we have two pos