A 0.696 mol sample of \(\mathrm{Cu}\) is added to \(136 \mathrm{mL}\) of \(6.0 \mathrm{M}\) HNO \(_{3}\). Assuming the following reaction is the only one that occurs, will the Cu react completely? $$\begin{aligned} 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) +4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{NO}(\mathrm{g}) \end{aligned}$$

First, the moles of HNO3 are calculated from the given volume and molarity, using the equation \( Molarity = Moles/Volume \). Here the volume should be converted to liters by dividing by 1000, so the calculation would be: \( Moles_{HNO3} = Molarity_{HNO3} * Volume_{HNO3}/1000 = 6M * 136ml/1000 = 0.816 mol \)

Next, it's necessary to determine which reactant is limiting, i.e., will run out first and stop the reaction. According to the balanced equation, 3 moles of copper are needed for 8 moles of nitric acid. But there are 0.696 mol of copper and 0.816 mol of nitric acid in the reaction. So, calculate how much nitric acid is needed for available copper and how much copper is needed for available nitric acid. This is found from the following: \( Needed HNO3 = 8/3 * Available Cu = 8/3 * 0.696 = 1.85 mol \). And, \( Needed Cu = 3/8 * Available HNO3 = 3/8 * 0.816 = 0.306 mol \). Comparing these, we see that the amount of needed nitric acid for the available copper is more than the provided, so nitric acid is the limiting reactant.

For the last step, determine if all copper reacts. From the calculations, the available nitric acid required 0.306 mol of copper for the reaction to complete. The total amount of copper available was 0.696 mol, which is much more than needed. Therefore, not all the copper will react completely.