\(n_{Al}=\frac{3,6}{27}=\frac{2}{15}\left(mol\right);n_{NaOH}=0,1.1=0,1\left(mol\right)\)
\(PTHH:Al+NaOH+H_2O\rightarrow NaAlO_2+\frac{3}{2}H_2\)
(mol)___________0,1______________________0,15__
Tỉ lệ: \(\frac{\frac{2}{15}}{1}>\frac{0,1}{1}\rightarrow Al\cdot du\)
\(V_{H_2}=22,4.0,15=3,36\left(l\right)\)
n Al=3,6/27=0,133(mol)
n NaOH=0,1.1=0,1(mol)
----> Al dư
Tính theo số mol của NaOH
Al+2NaOH+H2O--->NaAlO2+2H2
Theo pthh
n Al=1/2n NaOH=0,05(mol)
m Al=0,05.27=1,35(g)
Ý mk làm lộn
Al+2NaOH+H2O--->NaAlO2+2H2
n Al=3,6/27=0,133(mol)
n NaOH=0,1.1=0,1(mol)
--->Al dư
Theo pthh
n H2=n NaOH=0,1(mol)
V H2=0,1.22,4=2,24(l)
Ta có :
\(n_{Al}=0,133\left(mol\right)\)
\(n_{NaOH}=0,1\left(mol\right)\)
\(PTHH:Al+2NaOH+H2O\rightarrow NaAlO2+2H2\)
=>nH2 = n NaOH = 0,1(mol)
=>VH2 = 0,1 . 22,4 = 2,24l
nAl=3,6:27=0,13(mol)
nNaOH=1.0,1=0,1(mol)
PTHH:Al+NaOH+H2O->NaAlO2+\(\frac{3}{2}\)H2
mol:.....1............1......1.......1...........\(\frac{3}{2}\)
mol:.......0,1.......0,1..........................0,15
ta có tỉ lệ:\(\frac{0,13}{1}\)>\(\frac{0,1}{1}\)=>Al dư
V=0,15.22,4=3,36(l)
