stoichiometry ??

Question

6.56 grams of copper metal is added to concentrated nitric acid to yeld nitrogen momoxide, water, and copper(ii) nitrate. how many grams of nitric acid were used to fully react with the copper metal??

somebody help cause im really stuck.
thanks...

Answer 1

well the chemical equation for this problem is written in this form:

Cu + HNO3 -> NO + H2O + Cu(NO3)2

Why is the last term written like this? the phrase "copper (II)" means that the copper ion in the compound has a charge of +2. The nitrate ion NO3 has a charge of -1, so 2 nitrate ions are needed to balance the charge of the copper ion.

Now we need to balance the chemical equation to get the correct stoichiometry.

To find the values of these coefficients we need to do atomic balances of H, N, O, and Cu:

v Cu + w HNO3 -> x NO + y H2O + z Cu(NO3)2

The equation can be scaled to whatever quantity we want. So for now lets assume we have 100 moles of HNO3 in this equation, so w = 100

N balance: 100(1 N atom) = x(1 N atom) + z(2 N atoms)
100 = x + 2z

O balance:
100(3 atoms) = x + y + z(3atoms)(2 atoms)
300 = x + y + 6z

H balance:
100 = 2y

Cu balance:
v = z

so y = 50
x + 2z = 100
x + y + 6z = 300
x + 6z = 250
4z = 150 z = v = 37.5
x = 25

So one balanced equation is
37.5 Cu + 100 HNO3 -> 25 NO + 50 H2O + 37.5 Cu(NO3)2

A simpler equation may be determined by dividing by 12.5
3 Cu + 8 HNO3 -> 2 NO + 4 H2O + 3 Cu(NO3)2

Now we know that it takes 8 moles of HNO3 to react 3 moles of Cu metal

The molecular weight of Cu is 63.55 g/mol

So the number of copper moles:
6.56 g Cu (1 mol Cu/ 63.55 g Cu) = 0.10322 mol Cu

Moles of HNO3 needed to fully react with Cu
0.10322 mol Cu(8 mol HNO3/ 3 mol Cu) = 0.27527 mol HNO3

Mass of HNO3 needed
0.27527 mol HNO3(63.01 g HNO3/ 1 mol HNO3) = 17.3 grams HNO3.

Answer 2

26.0 grams of concentrated nitric acid is needed.

There is already a little problem with your question. Copper and concentrated nitric acid produces nitrogen DIOXIDE "NO2". Copper and dilute nitric acid produces nitrogen monoxide "NO".

First you need to write the balanced chemical reaction. For concentrated nitric acid, the equation is:

Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)


For dilute nitric acid the equation is:

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


"Balanced" means you have the same number of atoms of each element on the left side and right side. Always try to look up the reaction on the internet, these equations are from:

http://www.angelo.edu/faculty/kboudrea/demos/copper_HNO3/Cu_HNO3.htm

The (s) means solid, (aq) means aqueous ... dissolved in water, (g) means gas, and (l) means liquid.


Since it is most likely the typo is in the word "momoxide", only the case of concentrated nitric acid reacted with copper that produces nitrogen dioxide will be calculated. The equation is written below again for reference:

Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)


On the left side, notice that one atom of copper needs four molecules of HNO3 to fully react. The question only asked how much nitric acid was used, so whatever it reacts to on the right side can be ignored.

Next you need the atomic weights for copper and nitric acid.

Copper Cu = 63.543 grams/mole
Nitric Acid HNO3 = 63.01 grams/mole

Always look up atomic weights from a reference table.


Finally, calculate the number of moles of copper you have, and multiply by four. That is the number of moles of nitric acid you need to fully react. The number of moles of nitric acid times the molecular weight of nitric acid = the grams of nitric acid used.

6.56 grams copper divided by 63.543 grams/mole = 0.103 moles of copper.

0.103 x 4 = 0.413 moles of nitric acid needed.

0.413 moles x 63.01 grams/mole = 26.0 grams of nitric acid


For some reason, a lot of school chemistry problems come out to a round number in the end. If you get a round number, that is a good sign.

Answer 3

Well, you instructor has something wrong....Concentrated nitric acid forms nitrogen DIoxide. Weak nitric acid forms nitrogen MONOxide.

Assuming that the product is supposed to be MONOxide:

6.56 g / 63.55 g/mol = 0.103 mols

The reaction is:
3Cu(s) + 8HNO3(aq) ——> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
so 8 moles of nitric acid per 3 moles copper:

0.103 * (8/ 3) = 0.275 mols HNO3 or (0.275 + 63 g/mol) = 17.3g

Assuming nitrogen DIoxide the reaction is:
Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

And the ration is 4:1 so...
0.412 mols of HNO3 or 26.0g HNO3

Answer 4

what number moles of nitrogen and hydrogen are required to make one mole of ammonia, NH3? equation N2 + 3H2 = 2NH3 so dividing with the help of two to get one mole of NH3 provides a million/2 mole of N2 and 3/2 mole of H2