Question

there are reactions from 1 to 5.

Q: what is %yield of Cu?, and % error?

-reaction 1: 0.2134g of Cu, 2ml 6Mol HNO3

-reaction 5:0.1646g of Cu (solid) left.

Balanced Equation

reaction.1 Cu (s) + 4HNO3(aq) → Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l)

reaction 2: Cu(NO3)2 (aq) + 2NaOH (aq) → Cu(OH)2 (s) + 2NaNO3(aq)

reaction 3: Cu(OH)2 (s) → CuO (s) + H2O (l)

reaction 4: CuO (s) + H2SO4 (aq) → CuSO4 (aq) + H2O(l)

reaction 5: CuSO4 (aq) + Mg (s) → Cu (s) + MgSO4 (aq)

Answer 1

Here in the Counce of reaction from 1 ton, brinthiandig 1. Cu(s) + 4HNO3(aq) + Cu(NO3)2(aq) +2NO2 t2H2011) 2. Cu(NO3)2 609) +2 Naast (aq) → Cu(OH), (8) + 2NaNO3(aq) 3. CucoH), (5) Cuo(s) + H20 (1) 4. Cuo(s) + Hysoplag) Cuso (09) +Hole) 5. Cuso4 (29) + Mg(s) → Cu(s) + Mgso (ay) Initially, Imole of Cu(s) reacts with Amole of HNO3(aq) to give Imole of Cu(NO3)2, then in and reactions the genenated Linole of Cu(NO3) giver imole of Cuo(s). This produced Imole of Cho/s) reacts with Imole of H250, to produce Imóle Of Casoy can). finally the finale of cusog (aq) provide Imole of Culs) in reaction 5. :. In reaction I mass of Cu used = 0.21349 and 2ml_6(M) HNO3 = 2x6 > 12 mmole Hnog

According to neaction is Amole of HNO lof Hng, reacts Completely with a Imole lui mead Mass of the used foro heaction = 0.21349 meron - 72x10-3 mole aa a 125x12x10 millu Atomic mass of Cu: 63.59 g mol-' + 3x10-3 mde ce 2 3x10-363.54 g cu 2 19062x10-3 gcu "Mass of unneacted Cuin 0.1906 g cu Step1 oro reactions 2(0.234 -0.1906) g 20.02289 Cu Now, according to reaction 2,3,4,5 of stolcheom tamo neaction occurs in each step then we should get back 3x10-3 mole of ce am 0.1906 g cu in the final Step le, afters meaetima 5 .". Theomitical yield of Cu = 0.19069 But the given mass of Solid cu obtained in reaction to 2 0.16469 (practical yield) % geeld of Cu = yield obtained or practical yield (9) the omitical yield > 0116462_x100 , 86.388 0. 19069 4, emmarro (theomtical yield) (Practicalyieldx 100 themitical field 2 01906 – 011646 x 100 '2 13064 - 8100 0-1906 SEAT