Question

Answer 1

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

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

            0,3--->0,45-------->0,15-------->0,45

=> V_H2 = 0,45.22,4 = 10,08 (l)

b) m_muối = 0,15.133,5 = 20,025 (g)

c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,45}{0,2}=2,25M\)

Answer 2

Bài 3 :

a. \(n_{Al}=\dfrac{8.1}{27}=0,3\left(mol\right)\)

PTHH : 2Al + 6HCl -> 2AlCl₃ + 3H₂

              0,3     0,9        0,3         0,45

\(V_{H_2}=0,45.22,4=10,08\left(l\right)\)

b.  \(m_{AlCl_3}=133,5.0,3=40,05\left(g\right)\)

c. \(C_M=\dfrac{0.9}{0,2}=4,5\left(M\right)\)