`#AkaShi`
Câu `1` :
`a)`
PTHH:
`4Al + 3O_2` $\xrightarrow{t^o}$ `2Al_2O_3`
Theo đề: `m_{Y}=12,75=m_{Al_2O_3}`
`->n_{Al_2O_3}=(m_{Al_2O_3})/(M_{Al_2O_3})=(12,75)/(23*2+16*3)=0,125 (mol)`
`->n_{Al}=n_{Al_2O_3}xx4/2=0,125xx4/2=0,25 (mol)`
`->m_{Al}=n_{Al}xxM_{Al}=0,25xx27=6,75 (g)`
`b)`
PTHH:
`Al_2O_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2O`
Ta có: `n_{Al_2O_3}=0,125 (mol)`
`->n_{H_2SO_4}=n_{Al_2O_3}xx3=0,125xx3=0,375 (mol)`
`->m_{dd}=m_{Al_2O_3}+m_{H_2O_4}=12,75+80=92,75 (g)`
`->C%_{H_2SO_4}=(0,375xx98)/(92,75)xx100%=39,62%`
$\text{____________________________________}$`
Câu `2` :
PTHH: `2Na + 2H_2O -> 2NaOH + H_2`
`n_{Na}=(m_{Na})/(M_{Na})=(4,6)/23=0,2 (mol)`
`->n_{H_2}=1/2xxn_{Na}=0,2xx1/2=0,1 (mol)`
`->V_{H_2}=n_{H_2}xx22,4=0,1xx22,4=2,24 (l)`
`->n_{A}=n_{NaOH}=n_{Na}=0,2 (mol)`
`->C_{M(A)}=(n_{A})/(V_{A})=(0,2)/(0,5)=0,4 (M)`