If a 0.500 gram sample of copper metal reacts with 6 M nitric acid, what is the minimum volume (in mL) of acid should be added to ensure that all of the copper reacts? Show all work neatly

Question

If a 0.500 gram sample of copper metal reacts with 6 M nitric acid, what is the minimum volume (in mL) of acid should be added to ensure that all of the copper reacts? Show all work neatly

Answer 1

I'd like to add a third answer, based on the observation that the HNO₃ is 6 M which is generally considered "dilute" nitric acid, as opposed to "concentrated" nitric acid. Because nitric acid is a strong oxy acid, and is a pretty good oxidizing agent, it will engage in oxidation reduction reactions as shown below.

3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O(l) ... balanced equation

0.5 g Cu x 1 mol Cu / 63.55 g x 8 mol HNO3 / 3 mol Cu = 0.02098 mols HNO₃ required

(x L)(6 mol/L) = 0.02098 mols

x = 0.003497 L = 3.50 mls of 6 M HNO3 would be needed

Answer 2

First you would have to get an idea of how 0.500 g of Copper translates to mol of the substance of Nitric acid, HNO3. If we are assuming that this is Copper I because it did not say or indicate that this is Copper II it will have a +1 charge and it will react as follows:

Cu + HNO3 ==> CuNO3 + H

If you can see that each species in the reaction has either a -1 or +1 charge there is one unit on each side and they are all balanced with a coeffecient of 1 in front of each.

so if you take 0.500g of the copper and turn this into moles, you will have an idea of how many moles of nitric acid or HNO3 you will have:

0.500g * 1mol/63,546g = 0.0078683 mol

this is the moles of copper and since, according to the balanced equation the ratio of Cu to HNO3 is 1:1

1mol of Cu ==> 1mol of HNO3

0.0078683 mol Cu ===> 0.0078683 mol of HNO3

because it is a 1:1 ratio

so you now know you have 0.0078683 mol of HNO3

to figure out what this means for volume use the molarity given: 6 M

M is also equal to mol/L so,

M=mol/L

6M = 0.0078683mol/ L

Now you can solve for L or liters and you get:

0.0078683 mol = 6M*L

0.0078683 mol/6M = L = 0.001311383 L

Now to

1000ml = 1L

0.001311383 L * 1000mL/1 L so now Liters will cancel each other top to bottom and you will have mL or milliliters

So at least

1.31 mL should be added

Answer 3

The balanced equation is Cu(s) + 2 HNO3(aq) ====> Cu(NO3)2 (aq) + H2(g).

As 2 moles of HNO3 can react with 1 mole Cu (= 63.55 g), a conversion factor is 2 moles HNO3 / 63.55 g Cu. If we multiply this factor by 0.500 g Cu, we get 2 moles HNO3 / 63.55 g Cu x 0.500 g Cu = 0.0157 moles HNO3 required.

As the HNO3 is 6 M, this means there are 1000 mL / 6 moles HNO3 or 166.7 mL per mole HNO3.

Multiply 166.7 mL / mole HNO3 x 0.0157 moles HNO3 = 2.62 mL of 6M HNO3 required.