Hóa học Hòa tan hoàn toàn 10g hh Al, Al2O3 trong NaOh dư thu đc 6,72 líth2. %m Al = 17/11/2021 By Piper Hòa tan hoàn toàn 10g hh Al, Al2O3 trong NaOh dư thu đc 6,72 líth2. %m Al =
$2Al+ 2NaOH+ 2H_2O \rightarrow 2NaAlO_2+ 3H_2$ $n_{H_2}= 0,3 mol$ => $n_{Al}= 0,2 mol$ %Al= $\frac{0,2.27.100}{10}$= 54% Trả lời
Đáp án: \[n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\] \[2Al+2NaOH+6H_2O\to 2Na[Al(OH)_4]+3H_2\] \[0,2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leftarrow\quad \quad\quad0,3\qquad(mol)\] \[Al_2O_3+2NaOH+3H_2O\to 2Na[Al(OH)_4]\] \[\to m_{AL}=0,2\times 27=5,4(g)\] \[\to \%m_{Al}=\dfrac{5,4}{10}\times 100\%=54\%\] Trả lời
$2Al+ 2NaOH+ 2H_2O \rightarrow 2NaAlO_2+ 3H_2$
$n_{H_2}= 0,3 mol$
=> $n_{Al}= 0,2 mol$
%Al= $\frac{0,2.27.100}{10}$= 54%
Đáp án:
\[n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\]
\[2Al+2NaOH+6H_2O\to 2Na[Al(OH)_4]+3H_2\] \[0,2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leftarrow\quad \quad\quad0,3\qquad(mol)\]
\[Al_2O_3+2NaOH+3H_2O\to 2Na[Al(OH)_4]\]
\[\to m_{AL}=0,2\times 27=5,4(g)\]
\[\to \%m_{Al}=\dfrac{5,4}{10}\times 100\%=54\%\]