A student performing an experiment mistakenly used 8.0 ml of 16 M HNO3 to dissolve 0.16g of solid copper instead of the 4.0 ml as suggested. how many moles of HNO3 remains after the reaction with copper is complete. 8 moles of HNO3 reacts with 3 moles of Cu

im not sure how i would start this equation. would i start from the moles of HNO3 calcualted from the 8ml of 16 M HNO3 or with the .16 g of Cu

yes and yes.

8HNO3 + 3Cu ==>
mols HNO3 initially = M x L = ?
mols Cu = grams/atomic mass = ?

Now calculate how much HNO3 is need to dissolve that many mols Cu.

mols HNO3 remaining unreacted = initial mols HNO3 = mols needed to dissolve the Du = ?

To solve this problem, you need to determine the limiting reactant first. The limiting reactant is the one that gets fully consumed and determines the amount of product formed.

Let's start by calculating the number of moles of HNO3 used. We are given that the student mistakenly used 8.0 ml of 16 M HNO3. To calculate the moles of HNO3, we can use the following formula:

Moles of HNO3 = (volume of solution in liters) x (concentration of the solution in moles per liter)

Converting the volume from milliliters (ml) to liters (L), we have:

Volume of HNO3 solution = 8.0 ml = 8.0/1000 L = 0.008 L

Now, let's calculate the moles of HNO3 used:

Moles of HNO3 = 0.008 L x 16 mol/L = 0.128 mol of HNO3

Next, we need to determine the moles of copper (Cu) used. Given that the mass of copper used is 0.16 g, we can calculate the moles of Cu using its molar mass, which is 63.55 g/mol:

Moles of Cu = mass of Cu / molar mass of Cu

Moles of Cu = 0.16 g / 63.55 g/mol ≈ 0.0025 mol of Cu

Now, we know that the stoichiometric ratio of HNO3 to Cu is 8:3. This means that 8 moles of HNO3 react with 3 moles of Cu. Therefore, to find the moles of HNO3 remaining, we need to find the difference between the moles used and the stoichiometric amount based on the moles of Cu used.

Moles of HNO3 remaining = Moles of HNO3 used - Stoichiometric ratio x Moles of Cu used

Moles of HNO3 remaining ≈ 0.128 mol - (8/3) x 0.0025 mol

Moles of HNO3 remaining ≈ 0.128 mol - 0.0067 mol ≈ 0.1213 mol

Therefore, after the reaction with copper is complete, there would be approximately 0.1213 moles of HNO3 remaining.