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\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\) \(\Rightarrow y=0,03\left(mol\right)\)
\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)
\(n_{H_2}=\dfrac{0,448}{22,4}=0,02mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,02 0,02 ( mol )
\(\Rightarrow x=0,02\left(mol\right)\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{0,02}{0,03}=\dfrac{2}{3}\)
\(\Rightarrow CTHH:Fe_2O_3\)
\(n_{H_2\left(thu\right)}=\dfrac{V}{22,4}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
1 : 1 (mol)
0,02 : 0,02 (mol)
\(n_{H_2\left(dùng\right)}=\dfrac{V}{22,4}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
\(yH_2+Fe_xO_y\rightarrow^{t^0}xFe+yH_2O\)
y : x (mol)
0,03 : 0,02 (mol)
\(\Rightarrow\dfrac{0,03}{y}=\dfrac{0,02}{x}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{0,02}{0,03}=\dfrac{2}{3}\Rightarrow x=2;y=3\)
-Vậy CTHH của oxit sắt là Fe2O3.
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a+b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)
c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)
a,
nFe = 22,4/56 = 0,4 (mol)
PTHH
Fe2O3 + 3H2 ---to----) 2Fe + 3H2O (1)
theo phương trình (1) ,ta có:
nFe2O3 = 0,4 x 2 / 1 = 0,8 (mol)
mFe2O3 = 160 x 0,8 = 128 (g)
b,
theo pt (1)
nH2 = (0,4 x 3)/2 = 0,6 (mol)
=) VH2 = 0,6 x 22,4 = 13,44 (L)
c,
PTHH
Zn + H2SO4 -------------) ZnSO4 + H2 (2)
Số mol H2 cần dùng là 0,6 (mol)
Theo PT (2) :
nZn = nH2 ==) nZn = 0,6 x 65 = 39 (g)
\(\left(a\right)\)\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(\left(b\right)\)\(n_{Al}=\dfrac{4.05}{27}=0.15\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}=0.1125\left(mol\right)\Rightarrow V_{O_2}=2.52\left(l\right)\)
\(\Rightarrow n_{Al_2O_3}=\dfrac{n_{Al}}{2}=\dfrac{0.15}{2}=0.075\left(mol\right)\)
\(m_{Al_2O_3}=0.075\cdot102=7.65\left(g\right)\)
\(\left(c\right)\)
Để điều chế : 7.65 (g) Al2O3 thì cần 4.05 (g) Al và 2.52(l) khí O2
Vậy : để điều chế 25.5(g) Al2O3 thì cần x(g) Al và y(l) khí O2
\(m_{Al}=\dfrac{25.5\cdot4.05}{7.65}=13.5\left(g\right)\)
\(V_{O_2}=\dfrac{25.5\cdot2.52}{7.65}=8.4\left(l\right)\)
a) PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b) Ta có: \(n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\)
\(\Rightarrow n_{Al_2O_3}=0,075mol\) \(\Rightarrow m_{Al_2O_3}=0,075\cdot102=7,65\left(g\right)\)
c) Ta có: \(n_{Al_2O_3}=\dfrac{25,5}{102}=0,25\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,5mol\\n_{O_2}=0,375mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,5\cdot27=13,5\left(g\right)\\V_{O_2}=0,375\cdot22,4=8,4\left(l\right)\end{matrix}\right.\)
\(a.Đặt:\left\{{}\begin{matrix}Zn:x\left(mol\right)\\Mg:y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=0,3\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}65x+24y=11,3\\x+y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,1.65=6,5\left(g\right);m_{Mg}=0,2.24=4,8\left(g\right)\\ b.\%m_{Zn}=\dfrac{6,5}{11,3}=57,52\%;\%m_{Mg}=100-57,52=42,48\%\\ c.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O\\ TheoPT:n_{Fe_2O_3}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)
a)\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3mol\)
\(\rightarrow0,3molH_2\)\(\rightarrow V_{H2}=0,3.22,4=6,72l\)
b)\(H_2+Fe_2O_3\rightarrow3H_2O+2Fe\)
\(n_{Fe2O3}=\dfrac{19,2}{160}=0,12mol\)
\(\rightarrow0,24molFe\rightarrow m_{Fe}=0,24.56=13,44gam\)
a.b.\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,1 0,1 0,1 ( mol )
\(m_{H_2SO_4}=0,1.98=9,8g\)
c.\(FeO+H_2\rightarrow\left(t^o\right)Fe+H_2O\)
0,1 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,3
\(m_{Al}=0,2.27=5,4\left(g\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,1 0,3
\(m_{Fe_2O_3}=160.0,1=16\left(g\right)\)
\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2<---------------------------0,3
\(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
0,1<------0,3
\(\left\{{}\begin{matrix}b,m_{Al}=0,2.27=5,4\left(g\right)\\c,m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)