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nH2=3,92/22,4=0,175(mol)
ZnO+H2---->Zn+H2O
x_____x_____x___x
CuO+H2--->Cu+H2O
y____y______y___y
Hệ pt:
\(\left\{{}\begin{matrix}81x+80y=14,1\\x+y=0,175\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,075\end{matrix}\right.\)
=>m chất rắn=x=0,1.65+0,075.64=11,3(g)
=>m H2O=y=18(0,1+0,075)=3,15(g)
nH2=3,92/22,4=0,175(mol)
ZnO+H2---->Zn+H2O
x_____x_____x___x
CuO+H2--->Cu+H2O
y____y______y___y
Hệ pt:
\(\left\{{}\begin{matrix}81x+80y=14,1\\x+y=0,175\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,075\end{matrix}\right.\)
=>m chất rắn=x=0,1.65+0,075.64=11,3(g)
=>m H2O=y=18(0,1+0,075)=3,15(g)
a)
\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ FeO + H_2 \xrightarrow{t^o} Fe + H_2O\)
b)
\(n_{H_2} = n_{H_2O} = \dfrac{14,4}{18} = 0,8(mol)\\ \Rightarrow m = m_X + m_{H_2} - m_{H_2O} = 64 + 0,8.2 - 14,4 = 51,2(gam)\)
PT: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)
\(Fe_3O_4+4CO\underrightarrow{t^o}3Fe+4CO_2\)
Giả sử: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_3O_4}=y\left(mol\right)\end{matrix}\right.\)
⇒ 80x + 232y = 39,2 (1)
Ta có: \(n_{CO}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PT: \(n_{CO}=n_{CuO}+4n_{Fe_3O_4}=x+4y\left(mol\right)\)
⇒ x + 4y = 0,6 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
Bạn tham khảo nhé!
\(n_{CuO}=a\left(mol\right),n_{Fe_3O_4}=b\left(mol\right)\)
\(m_X=80a+232b=39.2\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
\(CuO+CO\underrightarrow{^{^{t^0}}}Cu+CO_2\)
\(Fe_3O_4+4CO\underrightarrow{^{^{t^0}}}3Fe+4CO_2\)
\(n_{H_2}=a+4b=0.6\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.1\)
\(\%Fe=\dfrac{0.1\cdot3\cdot56}{0.2\cdot64+0.1\cdot3\cdot56}\cdot100\%=56.75\%\)
H2+CuO->Cu+H2O
0,2---0,2-----0,2
Fe2O3+3H2-to>2Fe+3H2O
0,1-------0,3-------0,2
m CuO=32.\(\dfrac{50}{100}\)=16g
=>n CuO=\(\dfrac{16}{80}\)=0,2 mol
=>m Fe2O3=16g=>n Fe2O3=0,1 mol
=>m =mFe+m Cu=0,2.64+0,2.56=24g
c)Fe+H2SO4->FeSO4+H2
0,2---------------------0,2
=>m FeSO4=0,2.102=20,4g
Gọi x,y lần lượt là số mol CO2 , H2O
\(\left\{{}\begin{matrix}x+y=0,09\\44x+18y=\dfrac{40}{3}.2.0,09\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,03\\y=0,06\end{matrix}\right.\)
=> \(n_C=n_{CO_2}=0,03\left(mol\right);n_H=2n_{H_2O}=0,12\left(mol\right)\)
\(BTKL\Rightarrow m_X=0,96\left(g\right)\)
=>\(M_X=\dfrac{0,96}{0,03}=32\)
\(BTNT\left(O\right):n_{O\left(trongX\right)}=0,03.2+0,06-0,045.2=0,03\left(mol\right)\)
Gọi CT của X : CxHyOz
x : y : z =0,03 : 0,12 : 0,03 = 1:4:1
=> CTĐGN : (CH4O)n
Mà \(M_X=32n=32\)
=> n=1
=> CT của X : CH4O
\(Đặt:\left\{{}\begin{matrix}a=n_{CH_4}\\b=n_{C_2H_4}\end{matrix}\right.\left(a,b>0\right)\\ a.CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\\ C_2H_4+3O_2\underrightarrow{^{to}}2CO_2+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+28b=3\\a+2b=\dfrac{4,48}{22,4}=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ b.V_{O_2\left(đktc\right)}=22,4.\left(2a+3b\right)=22,4.\left(2.0,1+0,05.3\right)=7,84\left(l\right)\\ c.\%V_{CH_4}=\dfrac{a}{a+b}.100\%=\dfrac{0,1}{0,05+0,1}.100\approx66,667\%\\ \Rightarrow\%V_{C_2H_4}\approx33,333\%\)
nH2=3,92/22,4=0,175(mol)
ZnO+H2---->Zn+H2O
x_____x_____x___x
CuO+H2--->Cu+H2O
y____y______y___y
Hệ pt:
{81x+80y=14,1x+y=0,175⇒{x=0,1y=0,075{81x+80y=14,1x+y=0,175⇒{x=0,1y=0,075
=>m chất rắn=x=0,1.65+0,075.64=11,3(g)
=>m H2O=y=18(0,1+0,075)=3,15(g)