Zadanie 1b
Wzorowane na: Chemistry HL Paper 2, May 2026, TZ3, pytanie 1(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 1a-1c.
Metals combine with non-metals to form ionic compounds, many of which are used in industry and in the laboratory.
Polecenie
Potassium also forms compounds with polyatomic ions.
2 KOH(aq) + Cu(NO3)2(aq) → Cu(OH)2(s) + 2 KNO3(aq)
(i) 30.0 cm3 of a 0.380 mol dm−3 potassium hydroxide, KOH, solution is mixed with 25.0 cm3 of a 0.310 mol dm−3 copper(II) nitrate, Cu(NO3)2, solution. Determine the limiting reactant and the mass of copper(II) hydroxide, Cu(OH)2, produced, showing your working. Use sections 1 and 7 of the data booklet.
(ii) Calculate the percentage yield of copper(II) hydroxide from part (b)(i) if 0.472 g of the product was collected. If you did not obtain an answer in part (b)(i), use 0.650 g, although this is not the correct answer.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
(i) «n(KOH) = 0.0300 dm3 × 0.380 mol dm−3 =» 0.0114 «mol» AND «n(Cu(NO3)2) = 0.0250 × 0.310 =» 0.00775 «mol» ✔
«2:1 ratio, so 0.00775 mol of Cu(NO3)2 would need 0.0155 mol of KOH; only 0.0114 mol is available, hence» KOH is the limiting reactant ✔
«n(Cu(OH)2) = 0.0114/2 = 0.00570 mol; M = 97.57 g mol−1; m = 0.00570 × 97.57 =» 0.556 «g» ✔
(ii) «0.472/0.556 × 100 =» 84.9 «%» ✔
Schemat punktowania
(i) Award [3] for the correct final answer with working. M1: both amounts of reactant; M2: limiting reactant identified with a justification that uses the 2:1 ratio (do not award M2 for the bare statement); M3: mass of product from the limiting reactant. (ii) [1]; if 0.650 g is used the answer is 72.6 %. Ignore sig. figs and units unless clearly wrong.