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\(n_{AgNO_3}=0,1.0,2=0,02\left(mol\right)\\ n_{HCl}=0,3.0,2=0,06\left(mol\right)\\ AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ a,Vì:\dfrac{0,02}{1}< \dfrac{0,06}{1}\Rightarrow HCldư\\ \Rightarrow n_{AgCl}=n_{HNO_3}=n_{AgNO_3}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,02=2,87\left(g\right)\\ b,dd.sau.p.ứ:HNO_3,HCl\left(dư\right)\\ n_{HCl\left(dư\right)}=0,06-0,02=0,04\left(mol\right)\\V_{ddsau}=V_{ddAgNO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ \Rightarrow C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,4}=0,1\left(M\right)\\ C_{MddHNO_3}=\dfrac{0,02}{0,4}=0,05\left(M\right)\)
AgCl kết tủa thì không có phải dung dịch nên không tính nồng độ mol đâu bạn
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
a) Pt : \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl|\)
1 2 1 2
0,25 0,5 0,25
b) \(n_{Fe\left(OH\right)2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
⇒ \(m_{Fe\left(OH\right)2}=0,25.90=22,5\left(g\right)\)
c) \(n_{FeCl2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
100ml = 0,1l
\(C_{M_{ddFeCl2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
d) Pt : \(Fe\left(OH\right)_2\underrightarrow{t^o}FeO+H_2O|\)
1 1 1
0,25 0,25
\(n_{FeO}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
⇒ \(m_{FeO}=0,25.72=18\left(g\right)\)
Chúc bạn học tốt
\(n_{FeCl_2}=\dfrac{150\cdot12.7\%}{127}=0.15\left(mol\right)\)
\(n_{NaOH}=\dfrac{350\cdot4\%}{40}=0.35\left(mol\right)\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
\(0.15...........0.3................0.15............0.3\)
\(m_{Fe\left(OH\right)_3}=0.15\cdot90=13.5\left(g\right)\)
\(m_{dd}=150+350-13.5=486.5\left(g\right)\)
\(C\%_{NaCl}=\dfrac{0.3\cdot58.5}{486.5}\cdot100\%=3.61\%\)
\(C\%_{NaOH\left(dư\right)}=\dfrac{\left(0.35-0.3\right)\cdot40}{486.5}\cdot100\%=0.4\%\)
\(Fe\left(OH\right)_2\underrightarrow{^{^{t^0}}}FeO+H_2O\)
\(0.15..........0.15\)
\(m_{FeO}=0.15\cdot72=10.8\left(g\right)\)
\(4Fe\left(OH\right)_2+O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3+4H_2O\)
\(0.15.........................0.075\)
\(m_{Fe_2O_3}=0.075\cdot160=12\left(g\right)\)
\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,1______0,2_____0,1____0,1 (mol)
a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)
Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)
300ml = 0,3l
\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)
Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)
1 1 1 1
0,3 0,3 0,3
\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
200ml = 0,2l
\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)
Sau phản ứng :
\(V_{dd}=0,2+0,3=0,5\left(l\right)\)
\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
Chúc bạn học tốt
\(n_{HNO_3}=0,3\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)
\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)
a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,03---->0,03--------->0,03
=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)
b) mCH3COONa = 0,03.82 = 2,46 (g)
c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)
a) Đổi 300 ml = 0,3 l
Cm=\(\dfrac{n}{v}\) \(\Rightarrow\) nNaOH= Cm.V = 0,3.1 = 0,3 Mol
PTHH: 2NaOH + FeCl2 \(\rightarrow\) 2NaCl + Fe(OH)2
\(\Rightarrow\) nFe = 0,3 Mol
\(\Rightarrow\) VFe = \(\dfrac{n}{Cm}\) = \(\dfrac{0,3}{0,1}\) = 3 lít
b) ta có n\(_{Fe\left(OH\right)_2}\)= 0,3 Mol
\(\Rightarrow\) m\(_{Fe\left(OH\right)_2}\)= n.M = 0,3.90 = 27 g
c) mk nghĩ phải cho VNaCl mới giải đc
V = 300 ml = 0,3 (l)
=>nNaOH = 0,3.1 = 0,3 mol
2NaOH + FeCl2 ->2 NaCl + Fe(OH)2
0,3------>0,15---->0,3--------0,15
a) VFeCl2 = \(\dfrac{0,15}{0,1}\)= 1,5 (l)
b) mFe(OH)2 = 0,15 . 90 = 13,5 g
c) Vdd = 1,5+ 0,3 = 1,8 (l)
CM(NaCl) = \(\dfrac{0,3}{1,8}\) = 0,1666 M