Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
___0,2____________0,2_____0,3 (mol)
Có: m dd sau pư = mAl + m dd HCl - mH2 = 5,4 + 200 - 0,3.2 = 204,8 (g)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{204,8}.100\%\approx13,037\%\)
⇒ Đáp án: B
Bạn tham khảo nhé!