\(PTHH:Al(OH)_3+NaOH\to NaAlO_2+2H_2O\\ HCl+NaAlO_2+H_2O\to Al(OH)_3\downarrow+NaCl\\ n_{Al(OH)_3(\text{phản ứng})}=\dfrac{3,9}{78}=0,05(mol)\\ n_{NaOH}=3.0,05=0,15(mol)\\ n_{Al(OH)_3(\text{sản phẩm})}=\dfrac{1,56}{78}=0,02(mol)\)
Vì \(\dfrac{n_{Al(OH)_3}}{1}<\dfrac{n_{NaOH}}{1}\) nên \(NaOH\) dư
\(\Rightarrow n_{NaOH(dư)}=0,15-0,05=0,1(mol)\\ \Rightarrow n_{HCl}=n_{NaOH(dư)}+n_{Al(OH)_3}=0,02+0,1=0,12(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,12}{2}=0,06(l)\)
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