Đặt :
nCO = x mol
nH2 = y mol
nhh = x + y = 0.6 (1)
mhh = 28x + 2y = 0.5*30*0.6= 9 (2)
Giải (1) và (2) :
x = y = 0.3
Fe2O3 + 3CO -to-> 2Fe + 3CO2
________0.3______0.2
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
________0.3______0.2
mFe = 0.4*56=22.4 g
Vì : %V = %n
=> %H2 = %CO = 0.3/0.6*100% = 50%
Gọi x,y(mol) lần lượt là số mol CO và H2. ĐK: x,y>0.
Ta có: \(n_{hh}=\frac{13,44}{22,4}=0,6\left(mol\right)=x+y\left(1\right)\)
\(m_{hh}=0,5.M_{C_2H_6}.n_{hh}=9\left(g\right)\left(2\right)\)
Từ (1)(2)=>x=y=0,3(TM)
PTHH:
e2O3 + 3CO -----(to)---> 2Fe + 3CO2
...............0.3..........0.2.........................(mol)
Fe2O3 + 3H2 -----(to)----> 2Fe + 3H2O
.................0.3...........0.2.............................(mol)
K/l Fe:
mFe = 0.4*56=22.4 (g)
Vì : %V = %n
=> \(\%_{H_2}=\%_{CO}\) = \(\frac{0,3}{0,6}.100=50\left(\%\right)\)
\(m_{Fe_2O_3\left(1\right)}=x;n_{Fe_2O_3\left(2\right)}=y\)
\(PTHH_1:Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)
\(PTHH_2:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(n_{hh}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(M_{hh}=0,5.M_{C_2H_6}=15\)
\(\overline{M}=\frac{\sum_m}{\sum_n}\Rightarrow\sum_m=15.0,6=9\left(g\right)\)
Ta có: \(hpt:\left\{{}\begin{matrix}22,4\left(3x+3y\right)=13,44\\84x+6y=9\end{matrix}\right.\Leftrightarrow x=y=0,1\left(mol\right)\)
\(m_{Fe_2O_3}=160.\left(x+y\right)=160.\left(0,1+0,1\right)=32\left(g\right)\)
\(\%V_{CO}=\frac{22,4.3x}{13,44}.100\%=\frac{22,4.0,3}{13,44}.100\%=50\%\)
\(\%V_{H_2}=100-50=50\%\)
mFe2O3(1)=x ;nFe2O3(2)=y
PTHH1:Fe2O3+3COto→2Fe+3CO2
PTHH2:Fe2O3+3H2to→2Fe+3H2O
nhh=13,4422,4=0,6(mol)
Mhh=0,5.MC2H6=15
¯¯¯¯¯¯M=∑m∑n⇒∑m=15.0,6=9(g)
Ta có: hpt:{22,4(3x+3y)=13,44
84x+6y=9⇔x=y=0,1(mol)
mFe2O3=160.(x+y)=160.
%VCO=22,4*3x/13,44*100%=22,4*0,313,44*100=50%
%VH2=100−50=50%
Gọi x,y(mol) lần lượt là số mol CO và H2. ĐK: x,y>0.
Ta có: nhh=13,4422,4=0,6(mol)=x+y(1)
mhh=0,5.MC2H6.nhh=9(g)(2)
Từ (1)(2)=>x=y=0,3(TM)
PTHH:
e2O3 + 3CO -----(to)---> 2Fe + 3CO2
...............0.3..........0.2.........................(mol)
Fe2O3 + 3H2 -----(to)----> 2Fe + 3H2O
.................0.3...........0.2.............................(mol)
K/l Fe:
mFe = 0.4*56=22.4 (g)
Vì : %V = %n
=> %H2=%CO%H2=%CO = 0,30,6.100=50(%)
