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Ta có:
\(n_{Fe}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\Rightarrow n_{H2}=n_{Fe}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{8,4}{12}.100\%=70\%\\\%m_{Ag}=100\%-70\%=30\%\end{matrix}\right.\)
\(n_{HCl}=\dfrac{58,4.15\%}{36,5}=0,24\left(mol\right)\\ Fe+2HCl\rightarrow\left(t^o\right)FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,24}{2}=0,12\left(mol\right)\\ \Rightarrow V1=V_{H_2\left(đktc\right)}=0,12.22,4=2,688\left(l\right)\\ x=m_{Cu}=m_{hhA}-m_{Fe}=15,68-0,12.56=8,96\left(g\right)\\ b,n_{Cu}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,12+0,14=0,32\left(mol\right)\\ \Rightarrow V2=V_{Cl_2\left(đktc\right)}=0,32.22,4=7,168\left(l\right)\\ y=m_{muối}=m_{AlCl_3}+m_{CuCl_2}=0,12.133,5+0,14.135=34,92\left(g\right)\)
a, \(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)
\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\rightarrow m_{Al2O3}=15,8-5,6=10,2\left(g\right)\)
b, \(n_{HCl}=2n_{Fe}+6n_{Fe2O3}=0,8\left(mol\right)\)
\(\rightarrow m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(\rightarrow C\%_{HCl}=\frac{29,2}{50}.100\%=58,4\%=x\)
Đặt :
nFe = x mol
nMgO = y mol
mX = 56x + 40y = 13.6 (g) (1)
Fe + 2HCl => FeCl2 + H2
x____________x
MgO + 2HCl => MgCl2 + H2O
y______________y
mM = mFeCl2 + mMgCl2 = 127x + 95y = 31.7 (2)
(1) , (2) :
x = 0.1
y = 0.2
%Fe = 5.6/13.6 * 100% = 41.17%
%MgO = 58.82%
nKOH = 0.1 * 0.2 = 0.02 (mol)
KOH + HCl => KCl + H2O
0.02____0.02
nHCl (pư) = 2nFe + 2nMgO = 0.1*2 + 0.2*2 = 0.6 (mol)
nHCl = 0.02 + 0.6 = 0.62 (mol)
VddHCl = 0.62/0.5 = 1.24 (M)
Ta có:
\(\left\{{}\begin{matrix}n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\\n_S=\frac{4,8}{32}=0,15\left(mol\right)\end{matrix}\right.\)
\(PTHH:Fe+S\rightarrow FeS\)
\(\frac{0,1}{1}< \frac{0,15}{1}\) nên S dư
\(S+2HCl\rightarrow H_2S+Cl_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(CuSO_4+H_2S\rightarrow CuS+H_2SO_4\)
\(\left\{{}\begin{matrix}n_{H2S}=0,05+0,1=0,15\left(mol\right)\\n_{CuS}=n_{H2S}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{CuS}=0,15.96=14,4\left(g\right)\)
a)nfe=2.8/56=0.05mol
ns=2.4/32=0.075mol
fe +s-> fes
0.05 0.05 0.05
nfe<nS
->fe hết S dư,tính theo Fe
X là FeS và S
b)
mFeS=0.05*56=2.8g
mS.dư=(0.075-0.05)*32=0.8g