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\(m_{NaOH}=\dfrac{100\cdot10\%}{100\%}=10g\) \(\Rightarrow n_{NaOH}=0,25mol\)
\(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)
0,025 0,05 0,025
\(Zn\left(OH\right)_2\underrightarrow{t^o}ZnO+H_2O\)
0,025 0,025
\(m=m_{ZnO}=0,025\cdot\left(65+16\right)=2,025g\)
\(C_{M_{ZnCl_2}}=\dfrac{0,025}{\dfrac{500}{1000}}=0,05M\)
$n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)$
$Fe_3O_4 + 8HCl \to 2FeCl_3 + FeCl_2 + 4H_2O$
$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$
$FeCl_2 + 2NaOH \to Fe(OH)_2 + 2NaCl$
$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O$
Bảo toàn nguyên tố với Fe : $2Fe_3O_4 \to 3Fe_2O_3$
$n_{Fe_2O_3} = \dfrac{3}{2}n_{Fe_3O_4} = 0,015(mol)$
$m = 0,015.160 = 2,4(gam)$
\(n_{Fe}=\dfrac{6,5}{56}=\dfrac{13}{112}mol\)
\(m_{CH_3COOH}=\dfrac{90\cdot20\%}{100\%}=18g\Rightarrow n_{CH_3COOH}=0,3mol\)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\uparrow\)
\(\dfrac{13}{112}\) 0,3 0 0
\(\dfrac{13}{112}\) \(\dfrac{13}{56}\) \(\dfrac{13}{112}\) \(\dfrac{13}{112}\)
0 \(\dfrac{19}{280}\) \(\dfrac{13}{112}\) \(\dfrac{13}{112}\)
a)\(m_{\left(CH_3COO\right)_2Fe}=\dfrac{13}{112}\cdot174=20,2g\)
\(m_{H_2}=\dfrac{13}{112}\cdot2=\dfrac{13}{56}g\)
\(m_{dd\left(CH_3COO\right)_2Fe}=6,5+90-\dfrac{13}{56}=96,27g\)
\(C\%=\dfrac{20,2}{96,27}\cdot100\%=20,98\%\)
Mg + 2HCl -> MgCl2 + H2
0.2 0.4 0.2 0.2
\(nHCl=0.2\times2=0.4mol\)
a.\(m=0.2\times24=4.8g\); \(V=0.2\times22.4=4.48l\)
b.MgCl2 + 2NaOH -> Mg(OH)2 + NaCl
0.2 0.2
\(mNaOH=20\%\times100=20g\Rightarrow nNaOH=0.5mol\)
=> MgCl2 hết, NaOH dư
\(mMg\left(OH\right)2=0.2\times58=11.6g\)
m(ZnCl2)= 170*12/100=20,4g
n(ZnCl2)= 0,15mol
n(Zn(OH)2)=0,1mol < n(ZnCl2) =0,15
=> ZnCl2 dư
2NaOH + ZnCl2-> 2NaCl+Zn(OH)2
Số mol NaOH=2n(Zn(OH)2)=0,2 MOL
m(NaOH)= 8(g)
m(ddnaoh)=8*100/10=80(g)
\(n_{ZnCl_2}=\dfrac{170\cdot12\%}{136}=0.15\left(mol\right)\)
\(n_{Zn\left(OH\right)_2}=\dfrac{9.9}{99}=0.1\left(mol\right)\)
\(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)
TH1 : Kết tủa không bị hòa tan.
\(n_{NaOH}=2n_{Zn\left(OH\right)_2}=2\cdot0.1=0.2\left(mol\right)\)
\(m_{dd_{NaOH}}=\dfrac{0.2\cdot40}{10\%}=80\left(g\right)\)
TH2 : Kết tủa bị hòa tan một phần.
\(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)
\(0.15............0.3...........0.15\)
\(2NaOH+Zn\left(OH\right)_2\rightarrow Na_2ZnO_2+2H_2O\)
\(2x...........x\)
\(n_{Zn\left(OH\right)_2}=0.15-x=0.1\left(mol\right)\)
\(\Rightarrow x=0.05\)
\(n_{NaOH}=0.3+2\cdot0.05=0.4\left(mol\right)\)
\(m_{dd_{NaOH}}=\dfrac{0.4\cdot40}{10\%}=160\left(g\right)\)