Phản ứng xảy ra:
\(2Al+Fe_2O_3\underrightarrow{^{to}}2Fe+Al_2O_3\)
\(n_{Fe2O3}=\frac{8}{56.2+16.3}=0,05\left(mol\right)\)
Vì phản ứng xảy ra hoàn toàn mà khi cho NaOH vào X thu được khí.
Suy ra Al dư
\(NaOH+Al+H_2O\rightarrow NaAlO_2+\frac{3}{2}H_2\)
\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)
Ta có :
\(n_{Al\left(pư\right)}=2n_{Fe2O3}=0,1\left(mol\right),n_{H2}=0,15\left(mol\right)\)
\(\Rightarrow n_{Al\left(dư\right)}=0,1\left(mol\right)\)
\(n_{Al\left(bđ\right)}=0,1+0,15=0,25\left(mol\right)\)
\(\Rightarrow m=m_{Al}=0,25.27=6,75\left(g\right)\)
\(n_{NaAlO2}=n_{Al\left(bđ\right)}=0,25\left(mol\right)\)
\(NaAlO_2+CO_2+2H_2O\rightarrow NaHCO_3+Al\left(OH\right)_3\)
\(n_{Al\left(OH\right)3}=n_{NaAlO2}=0,25\left(mol\right)\)
\(\Rightarrow m_{Al\left(OH\right)3}=0,25.78=19,5\left(g\right)\)
