`#DyHungg`
`PTHH:` `Al_{2}O_{3} + 6HCl` `to` `2AlCl_{3} + 3H_{2}O`
b) `m_{HCl}=(C%xxm_{dd})/(100%)=(10%xx3,65)/(100%)=0,365(g)`
`n_{HCl}=(0,365)/(36,5)=0,01 (mol)`
`⇒n_{AlCl3}=(0,01xx2)/6=0,033(mol)`
`⇒m_{AlCl3}=0,033xx133,5=0,445(g0`
c) `⇒n_{Al2O3}=1/600 (mol)`
`⇒m_{Al2O3}=1/600xx102=0,17(g)`