Question

Answer 1

\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)

\(n_{HCl}=\dfrac{7,3}{36,5}=0,2mol\)

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

\(\dfrac{0,05}{2}\) < \(\dfrac{0,2}{6}\)                          ( mol )

 0,05     0,15                             ( mol )

Chất dư là HCl

\(m_{HCl\left(dư\right)}=\left(0,2-0,15\right).36,5=1,825g\)

Answer 2

2Al+6HCl->2AlCl3+3H2

0,05----0,15

n Al=\(\dfrac{1,35}{27}\)=0,05 mol

n HCl=\(\dfrac{7,3}{36,5}\)=0,2 mol

=>HCl dư

m HCl dư=0,05.36,5=1,825g