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a)
\(m_{CuO}=\dfrac{32.40}{100}=12,8\left(g\right)\) => \(n_{CuO}=\dfrac{12,8}{80}=0,16\left(mol\right)\)
\(n_{Fe_2O_3}=\dfrac{32-12,8}{160}=0,12\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,16->0,16---->0,16
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,12-->0,36----->0,24
=> \(V_{H_2}=\left(0,16+0,36\right).22,4=11,648\left(l\right)\)
b)
mCu = 0,16.64 =10,24 (g)
mFe = 0,24.56 = 13,44 (g)
c)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
Xét tỉ lệ: \(\dfrac{0,24}{1}< \dfrac{0,5}{2}\) => HCl dư, Fe hết
PTHH: Fe + 2HCl --> FeCl2 + H2
0,24------------------->0,24
=> \(V_{H_2}=0,24.22,4=5,376\left(l\right)\)
Đáp án:
8,96 l
Giải thích các bước giải:
a)
Fe2O3+3H2->2Fe+3H2O
CuO+H2->Cu+H2O
gọi a là số mol Fe2O3 b là số mol CuO
Ta có
160a=2x80b=>a=b
ta có
112a+64b=17,6
a=b
=>a=0,1 b=0,1
nH2=0,1x3+0,1=0,4(mol)
VH2=0,4x22,4=8,96 l
a)
CuO + H2 --to--> Cu + H2O
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
b) \(n_{Fe_2O_3}=\dfrac{32.20\%}{160}=0,04\left(mol\right)\)
\(n_{CuO}=\dfrac{32-0,04.160}{80}=0,32\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,32-->0,32---->0,32
Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,04-->0,12-------->0,08
=> VH2 = (0,32 + 0,12).22,4 = 9,856 (l)
c)
mCu = 0,32.64 = 20,48 (g)
mFe = 0,08.56 = 4,48 (g)
Bài 11:
\(a,n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\\ n_{Cu}=\dfrac{4}{80}=0,05\left(mol\right)\\ PTHH:\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
0,01 -----> 0,03 ---> 0,02
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,05 ---> 0,05 -> 0,05
\(b,m_{Fe}=0,02.56=1,12\left(g\right)\\ m_{Cu}=0,05.64=3,2\left(g\right)\\ V_{H_2}=\left(0,03+0,05\right).22,4=1,792\left(l\right)\)
Bài 12:
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{O\left(trong.oxit\right)}=n_{H_2}=0,15\left(mol\right)\\ n_{Fe\left(trong.oxit\right)}=\dfrac{8-0,15,16}{56}=0,1\left(mol\right)\\ CTHH:Fe_xO_y\\ \Rightarrow x:y=0,1:0,15=2:3\\ CTHH:Fe_2O_3\)
\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)
\(n_{Cu}=\dfrac{8}{80}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,02 0,06 0,04 ( mol )
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,1 0,1 0,1 ( mol )
\(m_{Fe}=0,04.56=2,24g\)
\(m_{Cu}=0,1.64=6,4g\)
\(n_{H_2}=0,06+0,1=0,16mol\)
\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)
\(m_{CuO}=\dfrac{8}{80}=0,1mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
\(CuO+H_2\rightarrow Cu+H_2O\)
\(m_{Fe}=0,02\cdot2\cdot56=2,24g\)
\(m_{Cu}=0,1\cdot64=6,4g\)
\(\Sigma n_{H_2}=0,02\cdot3+0,1=0,16mol\Rightarrow V_{H_2}=3,584l\)
a)
\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ b)\\ n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b =16,6(1)\\ n_{H_2} = 1,5a + b = \dfrac{11,2}{22,4} = 0,5(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,2\\ \Rightarrow m_{Al} = 0,2.27 = 5,4(gam)\ ;\ m_{Fe} = 0,2.56 = 11,2(gam)\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe_2O_3}=a\left(mol\right);n_{CuO}=b\left(mol\right)\left(a,b>0\right)\\ m_{hhoxit}=k\left(g\right)\\ \Rightarrow\left(1\right)160a+80b=k\\ \left(2\right)112a+64b=0,72k\\ \Rightarrow6,4a=12,8b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{12,8}{6,4}=\dfrac{2}{1}\\ \Rightarrow\%m_{Fe_2O_3}=\dfrac{160.2}{160.2+80.1}.100=80\%\\ \Rightarrow\%m_{CuO}=20\%\)
\(a) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ b) n_{CuO} = \dfrac{32.25\%}{80} = 0,1(mol)\\ n_{Fe_2O_3} = \dfrac{32-0,1.80}{160} = 0,15(mol)\\ n_{Cu} = n_{CuO} = 0,1(mol) \Rightarrow m_{Cu} = 0,1.64 = 6,4(gam)\\ n_{Fe} = 2n_{Fe_2O_3} = 0,3(mol) \Rightarrow m_{Fe} = 0,3.56 = 16,8(gam)\)