\(PTHH:A+2HCl\rightarrow ACl_2+H_2\)
pt:_______MA_________________22,4__
pứ:______9,75________________3,36__
Áp dụng ĐLTL:
\(\Rightarrow\frac{M_A}{9,75}=\frac{22,4}{3,36}\Leftrightarrow M_A=65\)
\(\rightarrow A:Zn\)
\(n_{HCl}=0,35.1=0,35\left(mol\right);n_{Zn}=\frac{9,75}{65}=0,15\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
(mol)____0,15___0,3______0,15_______
Lập tỉ lệ: \(\frac{0,15}{1}< \frac{0,35}{2}\rightarrow\) HCl dư
Số mol HCl dư là: \(n_{HCl\cdot du}=0,35-0,3=0,05\left(mol\right)\)
dd X lúc này là dd ZnCl2 và dd HCl dư
\(PTHH:ZnCl_2+2NaOH\rightarrow2NaCl+Zn\left(OH\right)_2\)
(mol)______0,15____0,3_________0,3__
\(PTHH:HCl+NaOH\rightarrow NaCl+H_2O\)
(mol)_____0,05____0,05_____0,05___
\(n_{NaOH}=0,4.1=0,4\left(mol\right)\)
Sau 2 pứ trên thì NaOH còn dư:
\(n_{NaOH\cdot du}=0,4-\left(0,05+0,3\right)=0,05\left(mol\right)\)
NaOH dư sẽ pứ tiếp với kết tủa Zn(OH)2
\(PTHH:Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\)
(mol)_______0,025_____0,05________0,025____
Vậy ddspứ gồm có \(NaCl;Na_2ZnO_2\)
\(C_{M_{NaCl}}=\frac{0,3+0,05}{0,35+0,4}=0,46\left(M\right)\)
\(C_{M_{Na_2ZnO_2}}=\frac{0,025}{0,35+0,4}=0,03\left(M\right)\)
\(\text{a)A+2HCl}\rightarrow\text{ACl2+H2}\)
Ta có :
\(n_A=n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(\Rightarrow M_A=\frac{9,75}{0,15}=65\left(g\right)\)
\(\rightarrow\)A là Zn
\(b,\text{nHCl=0,35.1=0,35(mol)}\)
\(\rightarrow\)nHCl dư =0,15(mol)
dd X gồm HCl dư và ZnCl2
\(\text{HCl+NaOH}\rightarrow\text{NaCl+H2O}\)
\(\text{ZnCl2+2NaOH}\rightarrow\text{Zn(OH)2+2NaCl}\)
\(\text{nNaOHdư=0,4-0,15.2-0,05=0,05(mol)}\)
\(\text{CMNaOH=}\frac{0,05}{0,75}=\frac{1}{15}M\)
\(\text{CMNaCl=}\frac{0,35}{0,75}=\frac{7}{15}M\)
