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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)
nSO2 = 3.36/22.4 = 0.15 (mol)
2Fe + 6H2SO4(đ) => Fe2(SO4)3 + 3SO2 + 6H2O
0.1...............................0.05.............0.15
mFe2O3 = 21.6 - 0.1*56 = 16 (g)
nFe2O3 = 16/160 = 0.1 (mol)
Fe2O3 + 3H2SO4 => Fe2(SO4)3 + 3H2O
0.1...................................0.1
mFe2(SO4)3 = ( 0.1 + 0.05) * 400 = 60 (g)
\(n_{H^+}=0,5.0,8+0,25.0,8.2=0,8\left(mol\right)\\ \Rightarrow n_{H_2}=\dfrac{n_{H^+}}{2}=\dfrac{0,8}{2}=0,4\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\)
Không phải KL hhX cho rồi à ta?
Bài 1:
\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
PTHH: Fe + 2HCl → FeCl2 + H2
\(m_{H_2}=0,16.2=0,32\left(g\right)\)
\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)
Bài 12:
Theo ĐLBTKL, ta có:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,05.56=2,8g\)
\(\Rightarrow m_{Zn}=0,35.65=22,75g\)
\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)
\(\%m_{Zn}=100\%-10,95\%=89,05\%\)
b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)
\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) (1)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\) (2)
Ta có: \(\Sigma n_{HCl}=1\cdot0,5=0,5\left(mol\right)\)
Gọi số mol của Fe là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\)
Gọi số mol của Fe2O3 là b \(\Rightarrow n_{HCl\left(2\right)}=6b\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}2a+6b=0,5\\56a+160b=13,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{FeCl_2}=0,1mol\\n_{FeCl_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\m_{FeCl_3}=0,1\cdot162,5=16,25\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{muối}=12,7+16,25=28,95\left(g\right)\)
Đặt :
nFe = x mol
nFe2O3 = y mol
mhh = 56x + 160y = 13.6 (g) (1)
Fe + 2HCl => FeCl2 + H2
x_____2x
Fe2O3 + 6HCl => 2FeCl3 + 3H2O
y________6y
nHCl = 2x + 6y = 0.5 (2)
(1) , (2) :
x = 0.1
y = 0.05
mFeCl2 = 0.1*127 = 12.7 (g)
mFeCl3 = 0.1*162.5 = 16.25 (g)