Đáp án:
Giải thích các bước giải:
`nAl = 0.81/27 = 0.03 mol`
`nS = 0.8/32 = 0.025 mol`
`2Al + 3S -to-> Al2S3`
Bđ: `0.03__0.025`
Pư:` 1/60__0.025____1/120`
Kt: `1/75___0_______1/120`
`2Al + 6HCl --> 2AlCl3 + 3H2`
`1/75__________________0.02`
`Al2S3 + 6HCl --> 2AlCl3 + 3H2S`
`1/120___________________0.025`
`V = ( 0.02 + 0.025 ) *22.4 = 1.008 (l)`
`mH2S = 0.85 g`
`mdd NaOH = 25*1.28 = 32 g`
`mNaOH = 32*15/100=4.8 g`
`nNaOH = 4.8/40 = 0.12 mol`
`NaOH/nH2S = 0.12/0.025=4.8 > 2`
=> Tạo ra muối Na2S, NaOH dư
2NaOH + H2S --> Na2S + 2H2O
0.05_____0.025___0.025
mdd sau phản ứng = 0.85 + 32 = 32.85 g
`mNa2S = 1.95 g`
`mNaOH dư = 2.8 g`
`C%Na2S = 5.93%`
`C%NaOH = 8.52%`