Đáp án đúng: C
Giải chi tiết:\(\begin{gathered} {n_{C{r_2}{O_3}}} = 0,01(mol) \hfill \\ \xrightarrow{{BTKL}}{m_{Al}} = {m_{chat\,ran}} - {m_{C{r_2}{O_3}}} = 2,33 - 1,52 = 0,81(g) \hfill \\ \to {n_{Al}} = 0,03(mol) \hfill \\ \,\,\,\,\,\,\,\,\,\,\,2Al + C{r_2}{O_3}\xrightarrow{{{t^o}}}A{l_2}{O_3} + 2Cr \hfill \\ Bd:0,03\,\,\,\,\,\,0,01 \hfill \\ Pu:0,02 \leftarrow 0,01 \to \,\,\,\,\,\,\,\,\,\,0,01 \to 0,02 \hfill \\ Sau:0,01\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,01\,\,\,\,\,\,\,\,0,02 \hfill \\ \xrightarrow{{BTCl}}{n_{HCl}} = 3{n_{AlC{l_3}}} + 2{n_{CrC{l_2}}} = 3.0,03 + 2.0,02 = 0,13(mol) \hfill \\ \to V = 1,3(l) \hfill \\ \end{gathered} \)
Đáp án C