The laboratory determines the content of titanium and iron in ilmenite containing only inert impurities using the following method:  

(1) Accurately weigh a 0.4278 g sample of ilmenite into a 30 mL porcelain crucible, add 10 g potassium pyrosulfate, and fuse at 700-750°C for decomposition. After cooling, leach the fused mass into a conical flask using 80 mL of 1:1 hydrochloric acid and 20 mL of 1:1 sulfuric acid in portions. Rinse the crucible with a small amount of hot water and transfer the washings to the same conical flask.  

(2) Add 3 g aluminum foil to the conical flask containing the leachate, and heat on an electric hotplate while slowly bubbling carbon dioxide until vigorous bubbling occurs. Cool to room temperature, add 10 mL of 400 g/L ammonium thiocyanate solution, and titrate with standard ferric ammonium sulfate solution until a blood-red color indicates the endpoint. The consumption of ferric ammonium sulfate solution is 25.49 mL.  

(3) Weigh 0.2174 g of high-purity titanium dioxide, dissolve it in acid, and treat it under the same two-step conditions described above. Titrate with standard ferric ammonium sulfate solution, consuming 26.79 mL.  

The correct options are (error less than 1% is considered correct):
实验室用如下方法测定仅含惰性杂质的钛铁矿中钛、铁的含量:

(1)准确称取钛铁矿样品$\mathrm{0.4278~g}$置于$\mathrm{30~mL}$瓷坩埚中,加入$\mathrm{10~g}$焦硫酸钾,在$\mathrm{700-750\degree C}$熔融分解。冷却后用$\mathrm{80~mL}$1:1盐酸和$\mathrm{20~mL}$1:1硫酸分次浸出熔块至锥形瓶中,
用少量热水洗涤坩埚将洗涤液转入同一锥形瓶中。

(2)向盛有浸取液的锥形瓶中加入$\mathrm{3~g}$铝箔,在二氧化碳缓缓鼓泡的条件下,在电热板上
加热溶解至冒大气泡。冷却至室温,加入$\mathrm{10~mL ~400~g/L}$的硫氰酸铵溶液,用硫酸铁铵标
准溶液滴定至血红色即为终点,消耗硫酸铁铵溶液$\mathrm{25.49~mL}$。

(3)称取高纯二氧化钛 $0.2174 \mathrm{~g}$ 溶于酸, 采用上述两步条件处理后用硫酸铁铵标准溶液滴
定, 消耗硫酸铁铵溶液$\mathrm{26.79~mL}$。

下列选项正确的是（误差小于1%即算正确）：
The titanium content in the mineral is $\mathrm{15.46\%}$.
The titanium content in the mineral is $\mathrm{24.83\%}$.
The titanium content in the mineral is $\mathrm{29.44\%}$.
The titanium content in the mineral is $\mathrm{35.75\%}$.
The iron content in the mineral is $\mathrm{20.99\%}$.
The iron content in the mineral is $\mathrm{26.30\%}$.
The iron content in the mineral is $\mathrm{35.88\%}$.
The iron content in the mineral is $\mathrm{40.22\%}$.
None of the above options is correct
矿物中钛元素含量为$\mathrm{15.46\%}$
矿物中钛元素含量为$\mathrm{24.83\%}$
矿物中钛元素含量为$\mathrm{29.44\%}$
矿物中钛元素含量为$\mathrm{35.75\%}$
矿物中铁元素含量为$\mathrm{20.99\%}$
矿物中铁元素含量为$\mathrm{26.30\%}$
矿物中铁元素含量为$\mathrm{35.88\%}$
矿物中铁元素含量为$\mathrm{40.22\%}$
以上选项均不正确
Ilmenite generally refers to $\mathrm{FeTiO_3}$; when fused with potassium pyrosulfate, it forms the highest-valence sulfates, namely $\mathrm{Fe_2(SO_4)_3,Ti(SO_4)_2}$.

<Checkpoint>[$\mathrm{FeTiO_3}$ forms $\mathrm{Fe_2(SO_4)_3,Ti(SO_4)_2}$ when fused with potassium pyrosulfate](1)</Checkpoint>

$\mathrm{Ti^{4+}}$ is unstable in solution and hydrolyzes to $\mathrm{TiO^{2+}}$:

$\mathrm{Ti^{4+} + H_2O = TiO^{2+} + 2H^+}$

<Checkpoint>[$\mathrm{Ti^{4+} + H_2O = TiO^{2+} + 2H^+}$](1)</Checkpoint>

$\mathrm{TiO^{2+}}$ is strongly oxidizing and can oxidize elemental aluminum:

$\mathrm{3TiO^{2+} + Al + 6H^+ = 3Ti^{3+} + Al^{3+} + 3H_2O}$

<Checkpoint>[$\mathrm{3TiO^{2+} + Al + 6H^+ = 3Ti^{3+} + Al^{3+} + 3H_2O}$](1)</Checkpoint>

The resulting $\mathrm{Ti^{3+}}$ is then converted back to $\mathrm{TiO^{2+}}$ by $\mathrm{Fe^{3+}}$ in the system:

$\mathrm{Ti^{3+} + Fe^{3+} + H_2O = TiO^{2+} + Fe^{2+} + 2H^+}$

<Checkpoint>[$\mathrm{Ti^{3+} + Fe^{3+} + H_2O = TiO^{2+} + Fe^{2+} + 2H^+}$](1)</Checkpoint>

Thus, it is equivalent to $\mathrm{Fe^{3+}}$ oxidizing aluminum first, followed by $\mathrm{TiO^{2+}}$ oxidizing aluminum.

<Checkpoint>[It is equivalent to $\mathrm{Fe^{3+}}$ oxidizing aluminum first, followed by $\mathrm{TiO^{2+}}$ oxidizing aluminum](1)</Checkpoint>

Excess aluminum is oxidized by $\mathrm{H^+}$.

<Checkpoint>[Excess aluminum is oxidized by $\mathrm{H^+}$.](1)</Checkpoint>

Ultimately, the system contains only $\mathrm{Fe^{2+}}$ and $\mathrm{Ti^{3+}}$, and ammonium ferric sulfate standard solution is added to titrate $\mathrm{Ti^{3+}}$ to the endpoint.

The standard solution of ammonium ferric sulfate is calibrated using pure titanium dioxide, with the reaction equation being the same as above and the stoichiometric ratio being 1:1, thus:

$\mathrm{c(Fe^{3+})=0.2174/79.87/0.002679=0.1016~mol/L}$

<Checkpoint>[The concentration of the ammonium ferric sulfate standard solution $\mathrm{c(Fe^{3+})=0.1016~mol/L}$](1)</Checkpoint>

The titanium content in the sample can be calculated as:

$\mathrm{\omega(Ti)=0.1016\times0.02549\times47.87/0.4278\times100\%=28.99\%}$

<Checkpoint>[$\mathrm{\omega(Ti)=28.99\%}$](1)</Checkpoint>

Based on the chemical formula of ilmenite and its inert impurities, the iron content is:

$\mathrm{\omega(Fe)=0.1016\times0.02549\times55.85/0.4278\times100\%=33.81\%}$

<Checkpoint>[$\mathrm{\omega(Fe)=33.81\%}$](1)</Checkpoint>

Consequently, options A-H are all incorrect, and option I is correct.
钛铁矿一般指$\mathrm{FeTiO_3}$；其在焦硫酸钾中共熔生成最高价硫酸盐，即$\mathrm{Fe_2(SO_4)_3,Ti(SO_4)_2}$。

<Checkpoint>[$\mathrm{FeTiO_3}$在焦硫酸钾中共熔生成$\mathrm{Fe_2(SO_4)_3,Ti(SO_4)_2}$](1)</Checkpoint>

$\mathrm{Ti^{4+}}$在溶液中不稳定，水解为$\mathrm{TiO^{2+}}$:

$\mathrm{Ti^{4+} + H_2O = TiO^{2+} + 2H^+}$

<Checkpoint>[$\mathrm{Ti^{4+} + H_2O = TiO^{2+} + 2H^+}$](1)</Checkpoint>

$\mathrm{TiO^{2+}}$具有强氧化性，可氧化铝单质：

$\mathrm{3TiO^{2+} + Al + 6H^+ = 3Ti^{3+} + Al^{3+} + 3H_2O}$

<Checkpoint>[$\mathrm{3TiO^{2+} + Al + 6H^+ = 3Ti^{3+} + Al^{3+} + 3H_2O}$](1)</Checkpoint>

此时生成的$\mathrm{Ti^{3+}}$会被体系中的$\mathrm{Fe^{3+}}$转化回$\mathrm{TiO^{2+}}$：

$\mathrm{Ti^{3+} + Fe^{3+} + H_2O = TiO^{2+} + Fe^{2+} + 2H^+}$

<Checkpoint>[$\mathrm{Ti^{3+} + Fe^{3+} + H_2O = TiO^{2+} + Fe^{2+} + 2H^+}$](1)</Checkpoint>

所以相当于是$\mathrm{Fe^{3+}}$先氧化铝，之后$\mathrm{TiO^{2+}}$再氧化铝。

<Checkpoint>[相当于是$\mathrm{Fe^{3+}}$先氧化铝，之后$\mathrm{TiO^{2+}}$再氧化铝](1)</Checkpoint>

过量的铝会被$\mathrm{H^+}$氧化。

<Checkpoint>[过量的铝会被$\mathrm{H^+}$氧化。](1)</Checkpoint>

最终体系只剩下$\mathrm{Fe^{2+}}$和$\mathrm{Ti^{3+}}$，加入硫酸铁铵标准溶液滴定$\mathrm{Ti^{3+}}$至终点。

硫酸铁铵的标准溶液通过纯二氧化钛标定，反应方程式与上面的方程式相同，化学计量比为1：1，从而

$\mathrm{c(Fe^{3+})=0.2174/79.87/0.002679=0.1016~mol/L}$

<Checkpoint>[硫酸铁铵的标准溶液浓度$\mathrm{c(Fe^{3+})=0.1016~mol/L}$](1)</Checkpoint>

可推算出样品中钛的含量

$\mathrm{\omega(Ti)=0.1016\times0.02549\times47.87/0.4278\times100\%=28.99\%}$

<Checkpoint>[$\mathrm{\omega(Ti)=28.99\%}$](1)</Checkpoint>

根据钛铁矿的化学式及其仅含惰性杂质可知铁的含量：

$\mathrm{\omega(Fe)=0.1016\times0.02549\times55.85/0.4278\times100\%=33.81\%}$

<Checkpoint>[$\mathrm{\omega(Fe)=33.81\%}$](1)</Checkpoint>

由此，选项A-H均错误，选项I正确。