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\(n_{CO_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH : \(C+O_2\underrightarrow{t^0}CO_2\)
PT : 1mol 1mol
Đề : 0,4mol ?mol
=> \(n_{O_2}=\frac{0,4\cdot1}{1}=0,4\left(mol\right)\)
=> \(V_{O_2}=0,4\cdot22,4=8,96\left(l\right)\)
\(V_{kk}\cdot20\%=V_{O_2}\Rightarrow V_{kk}=\frac{V_{O_2}}{20\%}=\frac{8,96}{20\%}=44,8\left(l\right)\)
=> \(V_{kk}=44,8l\)
a)\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,4 0,2 0,4
\(V_{O_2}=0,2\cdot22,4=4,48l\)
\(V_{kk}=5V_{O_2}=5\cdot4,48=22,4l\)
b)\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,4
\(m_{H_2O}=0,4\cdot18=7,2g\)
nCH4 = 4,48/22,4 = 0,2 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,2 ---> 0,4
Vkk = 0,4 . 22,4 : 21% = 128/3 (l)
\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15mol\)
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
0,15 0,3
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)
Không có đáp án đúng!
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(2H_2+O_2\rightarrow2H_2O\)
0,2 0,1 0,2 ( mol )
\(V_{kk}=\dfrac{V_{O_2}.100}{20}=\dfrac{\left(0,1.22,4\right).100}{20}=\dfrac{2,24.100}{20}=11,2l\)
\(m_{H_2O}=n_{H_2O}.M_{H_2O}=0,2.18=3,6g\)
nH2 = 4,48/22,4 = 0,2 (mol)
PTHH: 2H2 + O2 -> (t°) 2H2O
Mol: 0,2 ---> 0,1 ---> 0,2
Vkk = 0,1 . 22,4 . 5 = 11,2 (l)
mH2O = 0,2 . 18 = 3,6 (g)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,2.232=46,4\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\) \(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{21\%}\approx42,67\left(l\right)\)
d, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,4}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow n_{Fe_3O_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
a. \(3Fe+2O_2\rightarrow Fe_3O_4\)
b. Số mol Fe: \(n=\dfrac{m}{M}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
PTHH: \(3Fe+2O_2\rightarrow Fe_3O_4\)
Theo PTHH: \(3\) \(2\) \(1\) (mol)
Theo đề: \(0,6\) \(\rightarrow0,2\) (mol)
Kl của \(Fe_3O_4\) là: \(m=n\cdot M=0,2\cdot\left(56\cdot3+16\cdot4\right)=736\left(g\right)\)
a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4---------------->0,4
=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH4 dư, O2 hết
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4-------->0,2
=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
2H2+O2-to>2H2O
0,2----0,1-----0,2
n H2=0,2 mol
=>m H2O=0,2.18=3,6g
=>Vkk=0,1.22,4.5=11.2l
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 2H2 + O2 ----to----> 2H2O
Mol: 0,2 0,1 0,2
\(m_{H_2O}=0,2.18=3,6\left(g\right)\)
b, \(V_{O_2}=0,1.22,4=2,24\left(l\right)\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,4 0,2 ( mol )
\(V_{O_2}=0,2.22,4=4,48l\)
\(V_{kk}=V_{O_2}.5=4,48.5=22,4l\)