\(n_{Al} = \dfrac{5,4}{27} = 0,2(mol)\\ 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\\ n_{SO_2}= \dfrac{3}{2}n_{Al} = 0,3(mol) \Rightarrow V_{SO_2} = 0,3.22,4 = 6,72(lít)\\ n_{H_2SO_4} = 3n_{SO_2} = 0,6(mol)\\ \Rightarrow C_{M_{H_2SO_4}} = \dfrac{0,6}{0,04} = 15M\)
Đáp án A
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+6H_2SO_{4\left(đ\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(0.2........0.6....................................0.3\)
\(V_{SO_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.6}{0.04}=15\left(M\right)\)
PTHH: \(2Al+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,6\left(mol\right)\\n_{SO_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{H_2SO_4}}=\dfrac{0,6}{0,04}=15\left(M\right)\\V_{SO_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\) Chọn A
