`3)`

`1)`

`n_{Al}={2,7}/{27}=0,1(mol);n_{HCl}={73.25\%}/{36,5}=0,5(mol)`

`2Al+6HCl->2AlCl_3+3H_2`

`0,1->0,3->0,1->0,15(mol)`

`0,1<{0,5}/3->HCl` dư

`V_Y=V_{H_2}=0,15.22,4=3,36(L)`

`2)`

`m_{dd\ sau}=2,7+73-0,15.2=75,4(g)`

`C\%_{AlCl_3}={0,1.133,5}/{75,4}.100\%=17,71\%`

`C\%_{HCl\ du}={36,5(0,5-0,3)}/{75,4}.100\%=9,68\%`

`4)`

`CaCO_3` $\xrightarrow{t^o}$ `CaO+CO_2`

Theo pt:

`n_{CaCO_3\ t t}=n_{CaO\ t t}=7/{56}=0,125(tan\ mol)`

`n_{CaCO_3\ l t}={0,125}/{95\%}=0,13(tan\ mol)`

`m_{\text{quặng}}={0,13.100}/{80\%}=16,25(tan)`

`5)`

`1)`

`m_{NaOH\ bd}=120.20\%=24(g)`

`m_{NaOH\ them}=x(g)`

`->C\%_{NaOH\ sau}={24+x}/{120+x}.100=25`

`->x=8(g)`

`->m_{NaOH\ them}=8(g)`

`2)`

`85^oC:`

`S_{CuSO_4}={m_{CuSO_4}}/{m_{H_2O}}.100=87,7(g)`

`m_{CuSO_4}+m_{H_2O}=1877(g)`

`->m_{CuSO_4}=877(g);m_{H_2O}=1000(g)`

`12^oC:`

`n_{CuSO_{4}.5H_2O}=x(mol)`

`->n_{CuSO_4\ tach}=x(mol);n_{H_2O\ tach}=5x(mol)`

`->S_{CuSO_4}={877-160x}/{1000-18.5x}.100=35,5`

`->x=4,08(mol)`

`->m_{CuSO_{4}.5H_2O}=4,08.250=1020(g)`