3. (14%) The metal copper is oxidized in nitric acid according to the equation: 3 Cu + 8 HNO3 → 3 Cu (NO3) 2 + 2 NO + 4 H2O a) Show how you set the oxidation-reduction reaction and get the equation here above (show as many steps in the process) b) You receive 0.3855 g of copper-containing metal together with other dissolved substances easily set up. You find that to dissolve the copper and form copper nitrate Cu (NO3) 2 requires 22.5 ml of 0.575 M nitric acid. What is the ratio copper in the sample? Show how you calculate thx :*
Ans 3
Part a
The unbalanced reaction
Cu + HNO3 = Cu(NO3)2 + NO + H2O
Assign oxidation numbers for each atom within brackets
Cu(0) + H(+1)N(+5)O(-2)3 = Cu(+2)[N(+5)O(-2)3]2 + N(+2)O(-2) + H(+1)2O(-2)
The two half cell reactions are
Oxidation reaction
O
Cu(0) = Cu(+2)[N(+5)O(-2)3]2
Cu is being oxidized from 0 to +2
Reduction reaction
R:
H(+1)N(+5)O(-2)3 = N(+2)O(-2)
N is being reduced from 5 to 2
Balance the atoms except hydrogen and oxygen.
Oxidation reaction
Cu + 2HNO3 = Cu(NO3)2
Reduction reaction
HNO3 = NO
Balance the oxygen atoms by adding H2O
O:
Cu + 2HNO3 = Cu(NO3)2
R:
HNO3 = NO + 2H2O
Balance the hydrogen atoms by adding H+
O:
Cu + 2HNO3 = Cu(NO3)2 + 2H+
R:
HNO3 + 3H+ = NO + 2H2O
Balance the charge.
O:
Cu + 2HNO3 = Cu(NO3)2 + 2H+ + 2e-
R:
HNO3 + 3H+ + 3e- = NO + 2H2O
Make electron balance
O:
Cu + 2HNO3 = Cu(NO3)2 + 2H+ + 2e-
Multiply by 3
3Cu + 6HNO3 = 3Cu(NO3)2 + 6H++ 6e-
R:
HNO3 + 3H+ + 3e- = NO + 2H2O
Multiply by 2
2HNO3 + 6H+ + 6e- = 2NO + 4H2O
Add half reactions
3Cu + 8HNO3 + 6H+ + 6e- = 3Cu(NO3)2 + 2NO + 6H+ + 4H2O + 6e-
The balanced reaction
3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O
Part b
Mass of Cu = 0.3855 g
Moles of Cu = mass/molecular weight
= (0.3855g) / (63.546g/mol)
= 0.006066 mol
Volume of HNO3 = 22.5 mL x 1L/1000 mL = 0.0225 L
Molarity of HNO3 = 0.575 mol/L
Moles of HNO3 = molarity x volume
= 0.575 mol/L x 0.0225 L
= 0.0129375 mol
Mass of HNO3 = moles x molecular weight
= 0.0129375 mol x 63.01 g/mol
= 0.8152 g
Total mass of reactants = 0.3858 + 0.8152 = 1.201 g
Mass ratio of copper in the sample = mass of Cu/total mass
= 0.3858/1.201
= 0.3212
= 32.12%
3. (14%) The metal copper is oxidized in nitric acid according to the equation: 3 Cu...
please answer
Titration Homework 1. Copper reacts with dilute nitric acid according to the equation 3 Cu(s) + 8 HNO, (aq) + 3 Cu(NO3)2 (aq) + 2NO(g) + 4H20 (1) If a copper penny weighs 3.020g is dissolved in a small amount of nitric acid and the resulting solution is diluted to 50.0 mL with water, what is the molarity of the Cu(NO3)?
Copper reacts with nitric acid via the following equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(ℓ) What mass of NO(g) can be formed when 20.7 g of Cu reacts with 50.0 g HNO3? g NO
Metallic copper does not react with hydrochloric acid, but is oxidized by nitric acid since the nitrate ion functions as the oxidizing agent in acidic solutions. lf NO and Cu^2+ are formed as products, and the reaction is carried out under standard conditions, E^0 for the reaction of copper metal with nitric acid will be: +0.44 V +0.62 V +0.78 V +0.96 V HNO_3 (aq)+H_2O(l) NO_3^- (aq)+ H_3O^+ (aq) Cu(s)+ NO_3^- (aq) NO(g)+Cu^2+ (aq)
In lab, you reacted copper metal with aqueous nitric acid to produce aqueous Copper (11) nitrate, nitrogen dioxide gas and water. Cu (s) + 4HNO3 (aq) ---- > Cu(NO3)2 (aq) + 2NO2 (g) + 2H20 (1) a) If 0.210 gram of copper is reacted with 35.0 mL of 0.551 mol/L nitric acid, how many molecules and how grams of copper (II) nitrate are produced? b) How many moles of the reagent that is in excess are left over? c) If...
In lab, you reacted copper metal with aqueous nitric acid to produce aqueous copper (II) nitrate, nitrogen dioxide gas and water. Cu (s) + 4HNO3 (aq) ---- > Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l) a) If 0.210 gram of copper is reacted with 35.0 mL of 0.551 mol/L nitric acid, how many molecules and how grams of copper (II) nitrate are produced? b) How many moles of the reagent that is in excess are left over? c) If...
In lab, you reacted copper metal with aqueous nitric acid to produce aqueous copper (II) nitrate, nitrogen dioxide gas and water. Cu (s) + 4HNO3 (aq) ---- > Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l) a) If 0.210 gram of copper is reacted with 35.0 mL of 0.551 mol/L nitric acid, how many molecules and how grams of copper (II) nitrate are produced? b) How many moles of the reagent that is in excess are left over? c) If...
Copper start to finish lab questions 7a.) and 7b.)
hopefully the last 2 images of information help make the
problem clearer
7. a) For each mole of Cu dissolved in the nitric acid solution (HNO3), how many moles of [Cu(H2O)612+ are formed? col were formed in your reaction? b) Given your data above, how many moles of [Cu(H2O)6]2+ Reaction I:Cu (s)+ 4 H3O (aq) +2 NO3 (aq) --> [Cu(H2O)]** (aq) +2 NO2 (g) The first reaction in the series is...
Dissolution of Copper Metal Copper reacts readily with strong oxidizing agents (substances that readily remove electrons from other suhstances-i.e... Cu- Cu2 +2e). In this experiment, aqueous nitric acid, HNO, oxidizes copper metal to the copper(II) ion (opening photo): Cu(s) +4 HNO,(aq)Cu(NO)(aq) +2 NO (8) +2 H,0( .Cu HNO3 Cu(NO3)2 (28.1) CuSO The products of this reaction are copper(II) nitrate, Cu(NO)2 (a water-soluble salt that produces a blue solution), and nitrogen dioxide, NO2 (a dense, toxic, red-brown gas). The solution remains...
Write a balanced equation for the reaction of copper with nitric acxid tp produce nitric oxide (NO). Predict whether this reaction would occur with more, or less concentrated nitric acid than equation (1) : Cu(s) + HNO3(aq) ---> Cu(NO3)2(aq) + NO2(g) +H2O(l)
3a) Copper metal reacts with silver nitrate to form silver metal and copper nitrate according to the following equation; Balance the equation __Cu + __AgNO3 ---> __Ag + Cu(NO3)2 3b) You have 7.55 grams of each reactant, which is the limiting reagent? Show work to validate answer 3c)From the limiting reagent determined in 3b, calculate the grams of Ag that can be made.