Question

Answer 1

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,1}=1\left(M\right)\end{matrix}\right.\)