Question

Answer 1

\(n_{HCl}=\dfrac{10,95}{36,5}=0,3mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1         0,3         0,1         0,15  ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

Answer 2

2Al+6HCl->2Alcl3+3H2

0,1----0,3------0,1------0,15

n HCl=\(\dfrac{10,95}{36,5}\)=0,3 mol

=>m Al=27.0,1=2,7g

=>VH2=0,15.22,4=3,36l

m AlCl3=0,1.133,5=13,35g