IChO-2025_4.2-3
Type: Mechanistic Reasoning
Points: 21
Question: context: Chemists then discovered how to pressurise a ball above atmospheric pressure using a reaction between compounds A and B, which forms gas C (to create additional inner pressure) plus other non-gaseous products. A capsule with a quantity  m_{mix} = 0.3389 g of a stoichiometric mixture of two white inorganic compounds A and B was placed inside a ball filled with air at atmospheric pressure. They reacted completely inside the ball upon heating to form gas C and other products (reaction 1) to produce an overall pressure inside the ball of p = 1.600 atm (at 25°C). Once this ball was opened, all residue inside was collected and dissolved in water. The dropping of silver nitrate in that solution gave a white precipitate (reaction 2). This collected precipitate could be partially dissolved in a concentrated aqueous solution of B. By increasing the pH of this mixture, the precipitate dissolved completely (reaction 3). If the same initial capsule of A+B is dissolved in water and concentrated solution of silver nitrate added,  m_{precipitate} = 0.8224 g of white precipitate formed which was approximately twice the mass of precipitate formed in reaction 2. When the capsule of A + B dissolved in a solution of HI the formation of gas D started, and the solution turned brown (reaction 4). Upon further addition of sodium thiosulfate this colour disappeared (reaction 5). The densities of gases C and D differ by no more than  10%. Write the chemical formulae of compounds A-D. Write equations for reactions 1-5.
requirement: Compounds A-D must be identified through step-by-step reasoning with calculations and logical deductions. Simply stating the correct formulae without showing the reasoning process will not earn points. Reactions 1-5 must be written as complete, balanced chemical equations.
parsing_note: The reaction diagram has been converted to reaction_scheme. 'unknown' indicates unspecified reactants or products in the reaction.
reaction_scheme: reaction_number: 1
reactant_name: A + B
conditions: Δ
product_name: C, unknown
reaction_number: 2
reactant_name: unknown
conditions: AgNO_{3(aq)}
products: unknown
reaction_number: 3
reactant_name: unknown
conditions: B_{(aq)}, pH>7
products: unknown
reaction_number: 4
reactant_name: A + B
conditions: HI_{(aq)}
product_name: D, unknown
reaction_number: 5
reactant_name: unknow
conditions: Na_{2}S_{2}O_{3(aq)}
products: unknown
reactant_name: A + B
conditions: AgNO_{3(aq)}
products: unknown
Answer: step: 1
content: We can find the quantity of gas C: $n = \frac{pV}{RT} = \frac{(161975-101325) \text{ Pa} \times \frac{4}{3}\pi(0.03 \text{ m})^3}{8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 298.15 \text{ K}} = 2.767 \times 10^{-3} \text{ mol}$
points: 1
step: 2
content: The reaction with HI indicates a possible redox reaction with formation of $\text{I}_2$ (which is confirmed by the positive result in the further reaction with $\text{Na}_2\text{S}_2\text{O}_3$). Thus, A or B is an oxidant. If $\text{Ag}^+$ is added to A+B there is the formation of precipitate(s) (likely containing Ag(I)) but we don't know if it is one or several compounds. We note that reaction 2 gives approximately half the amount of precipitate as the final reaction. This could mean that the precipitate from the final reaction is a mixture of two precipitates, one of which is not formed in reaction 2.
points: 0
step: 3
content: We might first guess an equimolar amount of each precipitate from reaction 2 and that the number of moles of each precipitate is the same the number of moles of gas C, $(2.767 \times 10^{-3} \text{ mol})$. If this is the case, we can calculate a molar mass for the precipitate: $M = \frac{0.4112 \text{ g}}{2.767 \times 10^{-3} \text{ mol}} = 148.61 \text{ g mol}^{-1}$
points: 1
step: 4
content: Assuming the cation is Ag(I) (= 107.87 g mol$^{-1}$), this leaves an anion of 40.74 g mol$^{-1}$.
points: 1
step: 5
content: Remembering that this mass was only approximate, this is consistent with the anionic part being a chloride ion (= 35.45 g mol$^{-1}$) and the precipitate being AgCl. Upon addition of excess chloride, AgCl redissolves and thus this is consistent with B containing chloride ions.
points: 2
step: 6
content: Indeed, there is several qualitative indications that the precipitate could be AgCl. Now we know accurately the molar mass of the first precipitate, we can calculate the accurate molar mass of the second precipitate (= 153.90 g mol$^{-1}$). Assuming the cation is Ag(I), the anionic part has a molar mass of 46.03 g mol$^{-1}$.
points: 2
step: 7
content: There are no single atoms that fit this charge and mass combination. If we consider atoms of common elements then we can deduce the anion is nitrite (46.01 g mol$^{-1}$) and the precipitate as $\text{AgNO}_2$.
points: 2
step: 8
content: Hence, A is a nitrite salt and B is a chloride salt. Again, taking equimolar ratios as a first guess, we can calculate the sum of the cation molar masses in A+B, $(122.48 - 35.45 - 46.01)$ g mol$^{-1}$ = 41.02 g mol$^{-1}$.
points: 1
step: 9
content: As we know A is an oxidant due to the nitrite, reaction 1 is likely a redox reaction. However, it is the cation in B that is likely oxidised (as oxidation of chloride is difficult). One possibility would be the ammonium cation, and this also has a low molar mass.
points: 2
step: 10
content: The remaining molar mass $(41.02 - (14.01+4 \times 1.008))$ g mol$^{-1}$ = 22.98 g mol$^{-1}$ is consistent with the second cation being $\text{Na}^+$, so this is a possible answer.
points: 1
step: 11
content: $\text{NaNO}_2 + \text{NH}_4\text{Cl} \rightarrow \text{N}_2 + \text{NaCl} + 2\text{H}_2\text{O}$. We then conclude that C would be $\text{N}_2$.
points: 2
step: 12
content: If nitrite acts as an oxidant in reaction 4, then gas D has contains a nitrogen atom in an oxidation state of lower than +3. Examples include NO, $\text{N}_2\text{O}$ or $\text{NH}_3$, but only NO has a density similar to $\text{N}_2$.
points: 2
step: 13
content: The reaction 3 helps us understand that compound B is ammonium salt: $\text{NH}_4\text{Cl}$ thus A is $\text{NaNO}_2$
points: 0
reaction_number: 1
reactant_name: NaNO_{2} + NH_{4}Cl
reactant_formula: NaNO_{2} + NH_{4}Cl
conditions: Δ
product_name: N_{2} + NaCl + H_{2}O
product_formula: N_{2} + NaCl + 2H_{2}O
balanced_equation: $\text{NaNO}_{2} + \text{NH}_{4}\text{Cl} \rightarrow \text{N}_{2} + \text{NaCl} + 2\text{H}_{2}\text{O}$
points: 1
grading: partial_credit: condition: If not balanced correctly
deduction: -0.5
reaction_number: 2
reactant_name: Ag^{+} + Cl^{-}
reactant_formula: Ag^{+} + Cl^{-}
conditions: AgNO_{3(aq)}
product_name: AgCl
product_formula: AgCl
balanced_equation: $\text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl}$
points: 1
grading: partial_credit: condition: If not balanced correctly
deduction: -0.5
reaction_number: 3
reactant_name: AgCl + NH_{4}^{+} + OH^{-}
reactant_formula: AgCl + 2NH_{4}^{+} + 2OH^{-}
conditions: pH>7
product_name: Ag(NH_{3})_{2}^{+} + H_{2}O + Cl^{-}
product_formula: Ag(NH_{3})_{2}^{+} + 2H_{2}O + Cl^{-}
balanced_equation: $\text{AgCl} + 2\text{NH}_{4}^{+} + 2\text{OH}^{-} \rightarrow \text{Ag}(\text{NH}_{3})_{2}^{+} + 2\text{H}_{2}\text{O} + \text{Cl}^{-}$
points: 1
grading: partial_credit: condition: If not balanced correctly
deduction: -0.5
reaction_number: 4
reactant_name: NaNO_{2} + HI
reactant_formula: 2NaNO_{2} + 4HI
conditions: HI_{(aq)}
product_name: NO + I_{2} + NaI + H_{2}O
product_formula: 2NO + I_{2} + 2NaI + 2H_{2}O
balanced_equation: $2\text{NaNO}_{2} + 4\text{HI} \rightarrow 2\text{NO} + \text{I}_{2} + 2\text{NaI} + 2\text{H}_{2}\text{O}$
points: 1
grading: partial_credit: condition: If not balanced correctly
deduction: -0.5
reaction_number: 5
reactant_name: Na_{2}S_{2}O_{3} + I_{2}
reactant_formula: 2Na_{2}S_{2}O_{3} + I_{2}
conditions: Na_{2}S_{2}O_{3(aq)}
product_name: Na_{2}S_{4}O_{6} + NaI
product_formula: Na_{2}S_{4}O_{6} + 2NaI
balanced_equation: $2\text{Na}_{2}\text{S}_{2}\text{O}_{3} + \text{I}_{2} \rightarrow \text{Na}_{2}\text{S}_{4}\text{O}_{6} + 2\text{NaI}$
points: 1
grading: partial_credit: condition: If not balanced correctly
deduction: -0.5