Question

Answer 1

a) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right);n_{H_2SO_4}=0,2.2,5=0,5\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,5}{3}\) => H₂SO₄ dư, Al hết, tính theo Al

Theo PTHH: \(n_{H_2SO_4\left(pư\right)}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\)

=> \(V_A=V_{H_2}=0,45.24,79=11,1555\left(l\right)\)

b) Theo PTHH: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,15\left(mol\right)\)

=> \(m_{muối}=m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(n_{H_2SO_4\left(dư\right)}=0,5-0,45=0,05\left(mol\right)\)

=> \(\left\{{}\begin{matrix}C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,15}{0,2}=0,75M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,05}{0,2}=0,25M\end{matrix}\right.\)