What is the amount of nitric acid required to oxidize a #0.1025*g# mass of copper metal to #Cu(NO_3)_2# if its VOLUME is #200*mL#?

Question

Luna Chen · Accepted Answer

#"0.008065 mol L"^(-1)#

You must first write a balanced chemical equation for this redox reaction before moving further.

Nitric acid, #"HNO"_3#, is a very powerful oxidizing agent that will oxidize copper metal to copper(II) cations ,#"Cu"^(2+)#, while being reduced to nitrogen dioxide, #"NO"_2#, in the process.

Mind you, concentrated nitric acid will be reduced to nitrogen dioxide; by comparison, dilute nitric acid will be reduced to nitric oxide, #"NO"#.

Thus, this is how the chemical equation in balance appears.

#"Cu"_ ((s)) + 4"HNO"_ (3(aq)) -> "Cu"("NO"_ 3)_ (color(red)(2)(aq)) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O"_((l))#

Since nitric acid is a strong acid that completely dissociates in aqueous solution, you can rewrite the equation as follows: copper(II) nitrate is soluble in aqueous solution, meaning that it exists as cations and ions in solution.

#"Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 4"NO"_ (3(aq))^(-) -> "Cu"_ ((aq))^(2+) + color(red)(2)"NO"_ (3(aq))^(-) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O"_((l))#

You will then have the reaction's net ionic equation.

#"Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-) -> "Cu"_ ((aq))^(2+) + 2"NO"_ (2(g)) uarr + 2"H"_ 2"O"_((l))#

It's reasonable to assume that there will be an excess of acid present and that every mole of copper metal will participate in the reaction since you're working with concentrated nitric acid.

Use copper metal's molar mass to determine how many moles you have in that #"0.1025-g"# sample

#0.1025 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "0.001613 moles Cu"#

Since you have a #1:1# mole ratio between copper metal and copper(II) cations, you can say that the reaction will produce #0.001613# moles of copper(II) cations.

The total volume of the solution is said to be equal to #"200.0 mL"#. This means that the molarity of the copper(II)cations, which tells you how many moles of copper(II) cations you'd get per liter of this solution, will be equal to

#["Cu"^(2+)] = "0.001613 moles"/(200.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.008065 mol L"^(-1)color(white)(a/a)|)))#

I'll round the result to four significant figures.

Olivia Anton · Answer

undefined To determine the amount of nitric acid required to oxidize a given mass of copper metal to Cu(NO3)2, you need to first calculate the number of moles of copper present, then use the stoichiometry of the reaction between copper and nitric acid to find the amount of nitric acid needed. The balanced chemical equation for the reaction is:

3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

From the balanced equation, 3 moles of Cu reacts with 8 moles of HNO3. You'll need to use the molar mass of Cu and the volume of the solution to calculate the moles of Cu. Then, use stoichiometry to find the moles of HNO3 needed.

Given:
- Mass of Cu = 0.1025 g
- Volume of solution = 200 mL

1. Calculate moles of Cu:
   Moles = Mass / Molar mass

2. Use stoichiometry to find moles of HNO3 required using the ratio of Cu to HNO3 from the balanced equation.

3. Convert moles of HNO3 to grams using its molar mass if needed.