Zadanie 1a
Wzorowane na: Chemistry HL Paper 2, May 2024, TZ1, pytanie 1(a). 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-1e.
An effervescent powder for a soft drink has the following composition by mass:
| Component | Percentage by mass |
|---|---|
| Sucrose, C12H22O11 | 20.0 |
| Tartaric acid, C4H6O6 | 33.0 |
| Sodium hydrogencarbonate, NaHCO3 | 47.0 |
When water is added, the tartaric acid reacts with the sodium hydrogencarbonate and the powder fizzes:
2 NaHCO3(s) + C4H6O6(aq) → Na2C4H4O6(aq) + 2 CO2(g) + 2 H2O(l)
Polecenie
(i) Determine the limiting reactant when 2.00 g of this powder reacts.
(ii) Determine the volume, in dm3, of carbon dioxide released in the reaction in (i) at SATP. Use sections 1, 2 and 4 of the data booklet.
(iii) A student collects 0.190 dm3 of carbon dioxide from 2.00 g of the powder. Calculate the percentage yield of carbon dioxide.
If you did not obtain an answer to (ii), use 0.260 dm3, but this is not the correct value.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) M(C4H6O6) = 150.10 g mol−1 AND n(C4H6O6) = 0.660/150.10 = 0.00440 «mol» ✔
M(NaHCO3) = 84.01 g mol−1 AND n(NaHCO3) = 0.940/84.01 = 0.0112 «mol» ✔
«the equation needs 2 mol of NaHCO3 per mol of acid: 0.00440 × 2 = 0.00880 mol, which is less than 0.0112 mol, so» tartaric acid is the limiting reactant ✔
(ii) n(CO2) = 2 × 0.00440 = 0.00879 «mol» ✔
V = nRT/p = (0.00879 × 8.31 × 298.15)/100 = 0.218 «dm3» ✔
(iii) «0.190/0.218 × 100 =» 87.2 «%» ✔
Schemat punktowania
(i) Award [1] for either correct amount (or either correct molar mass). M3 only if the conclusion follows from the amounts calculated and the 1:2 ratio; accept the comparison made in moles of acid or in moles of hydrogencarbonate. (ii) M1: n(CO2) = 2n(C4H6O6), M2: volume; accept 0.217 to 0.218 dm3 (298 K or 298.15 K). ECF from (i): if sodium hydrogencarbonate was chosen as limiting, n(CO2) = 0.0112 mol and V = 0.277 dm3. (iii) [1]; accept 87 % (87.0 to 87.3 %); if 0.260 dm3 is used, 73.1 % (accept 73 %).