`#AkaShi`
PTHH: `6NaOH + Fe_2(SO_4)_3 -> 2Fe(OH)_3↓ + 3Na_2SO_4`
`->m_{NaOH}=40xx8%=8 (g)`
`->n_{NaOH}=8/40=0,2 (mol)`
`->m_{ Fe_2(SO_4)_3}=600xx10%=60 (g)`
`->n_{Fe_2(SO_4)_3}=60/400=0,15 (mol)`
Xét hết dư:
`n_{ Fe_2(SO_4)_3}=0,15 > n_{NaOH}=(0,2)/6 (mol)`
`->NaOH` hết ; ` Fe_2(SO_4)_3` dư
`->n_{Fe(OH)_3↓}=1/3xxn_{NaOH}=0,2xx1/3=1/15 (mol)`
`->m_{Fe(OH)_3↓}=1/15xx107=7,13 (g)`
`b)`
`->n_{Na_2SO_4}=0,2xx3/6=0,1 (mol)`
`->n_{Fe_2(SO_4)_3\ pư}=0,2xx1/6=1/30 (mol)`
`->n_{Fe_2(SO_4)_3\ dư}=0,15-1/30=7/60 (mol)`
`->m_{dd\ sau\ pư}=m_{NaOH}+m_{Fe_2(SO_4)_3}-m_{Fe(OH)_3↓}`
`=40+600-7,13=632,87 (g)`
`->C%_{Na_2SO_4}=(0,1xx142)/(632,87)xx100%=2,24%`
`->C%_{Fe_2(SO_4)_3\ dư}=(7/60xx400)/(632,87)xx100%=7,4%`
