n_{Al_2O_3}= {10,6}/{102}≈0,1(mol)

n_{HCl}= { 500.7,3%}/{36,5}=1(mol)

a)

$$\rm \ PTHH: Al_2O_3+6HCl → 2AlCl_3 + 3H_2O$$

(mol) 0,1→0,6→0,2→0,3

b)

Theo pt, vì: {0,1}/1 < 1/6 ⇒ HCl dư

→ Vì thế dd sau pứ làm quỳ tím hóa đỏ

c)

m_{HCl (dư)}= (1-0,6).36,5=14,6(g)

m_{AlCl_3} = 0,2.133,5=26,7(g)

a)\ 2Al+6HCl->2AlCl_3+3H_2

b)

n_{Al_2O_3}={10,6}/{102}\approx 0,1(mol)

n_{HCl}={500.7,3\%}/{36,5}=1(mol)

Do 0,1<1/6->HCl dư.

Do đó quỳ tím hóa đỏ.

c)

Theo PT: n_{AlCl_3}=2n_{Al_2O_3}=0,2(mol)

->m_{AlCl_3}=0,2.133,5=26,7(g)

Theo PT: n_{HCl\ pu}=3n_{AlCl_3}=0,6(mol)

->m_{HCl\ du}=(1-0,6).36,5=14,6(g)