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\(a,PTHH:Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol);n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2(mol)\)
Vì \(\dfrac{n_{H_2}}{4}<\dfrac{n_{Fe_3O_4}}{1}\) nên \(Fe_3O_4\) dư
\(n_{Fe_3O_4(dư)}=0,2-\dfrac{0,3}{4}=0,125(mol)\\ \Rightarrow m_{Fe_3O_4(dư)}=0,125.232=29(g)\\ b,n_{Fe}=\dfrac{3}{4}n_{H_2}=0,225(mol)\\ \Rightarrow m_{Fe}=0,225.56=12,6(g)\)
a) Fe3O4 + 4H2 --to--> 3Fe + 4H2O
b)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{4}\) => H2 hết, Fe3O4 dư
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,025<--0,1------>0,075
=> \(m_{Fe_3O_4\left(dư\right)}=\left(0,1-0,025\right).232=17,4\left(g\right)\)
c) \(m_{Fe}=0,075.56=4,2\left(g\right)\)
a)
\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)
b)
Ta có :
\(n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ n_{O_2} = \dfrac{96}{32} = 3(mol)\)
Ta thấy : \(\dfrac{n_{Fe}}{3} = 0,05 < \dfrac{n_{O_2}}{2} = 1,5\) do đó O2 dư.
Theo PTHH :
\(n_{O_2\ pư} = \dfrac{2}{3}n_{Fe} = 0,1(mol)\\ \Rightarrow n_{O_2\ dư} = 3 - 0,1 = 2,9(mol)\\ \Rightarrow m_{O_2\ dư} = 92,8(gam)\)
c)
\(n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = 0,05(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,05.232 = 11,6(gam)\)
\(a)PTHH:FeCl_3+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
mol 1 2 1
mol
\(b)\)Số mol \(FeCl_3\) là: \(n_{FeCl_3}=\dfrac{m_{FeCl_3}}{M_{FeCl_3}}=\dfrac{8,4}{162,5}=0,052\left(mol\right)\)
Số mol \(O_2\) là: \(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{96}{32}=3\left(mol\right)\)
Lập tỉ lệ: \(\dfrac{1}{0,052}>\dfrac{2}{3}\Rightarrow FeCl_3dư\)
Số mol \(FeCl_3\) phản ứng là:
Từ PTHH\(\Rightarrow\) \(n_{FeCl_3}=\dfrac{0,052\times3}{3}=0,035\left(mol\right)\)
Số mol \(FeCl_3\) dư là: \(n_{FeCl_3dư}=n_{FeCl_3đầu}-n_{FeCl_3p/ứng}=0,052-0,035=0,018\left(mol\right)\)
Khối lượng \(FeCl_3\) dư là: \(m_{FeCl_3dư}=n_{FeCl_3dư}\times M_{FeCl_3}=0,018\times162,5=2,925\left(g\right)\)
$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b) $n_{CH_4} = \dfrac{3,92}{22,4} = 0,175(mol)$
$n_{O_2} = \dfrac{3,84}{32} = 0,12(mol)$
Ta thấy : $n_{CH_4} : 1 > n_{O_2} : 2$ nên $CH_4$ dư
$n_{CH_4\ pư} = \dfrac{1}{2}n_{O_2} = 0,06(mol)$
$\Rightarrow m_{CH_4\ dư} = (0,175 - 0,06).16 = 1,84(gam)$
c) $2NaOH + CO_2 \to Na_2CO_3 + H_2O$
Theo PTHH :
$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$
`FeO + H_2` $\xrightarrow[]{t^o}$ `Fe + H_2 O`
`a) n_[H_2] = [ 3,36 ] / [ 22,4 ] = 0,15 (mol)`
`n_[FeO] = [ 14,2 ] / 72 = 71 / 360`
Ta có: `[ 0,15 ] / 1 < [ 71 / 360 ] / 1`
`=> FeO` dư
Theo `PTHH` có: `n_[FeO_\text{(p/ứ)}] = n_[H_2] = 0,15 (mol)`
`=> n_[FeO_\text{(dư)}] = 71 / 360 - 0,15 = 17 / 360 (mol)`
_______________________________________________
`b)` Theo `PTHH` có: `n_[Fe] = n_[H_2] = 0,15 (mol)`
`=> m_[Fe] = 0,15 . 56 = 8,4 (g)`
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)
\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{49}{98}=0,5\left(mol\right)\\ PTHH:Fe+H_2SO_4->FeSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) 0,4------->0,4-------->0,4-------->0,4
\(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\left(\dfrac{0,4}{1}< \dfrac{0,5}{1}\right)\)
`=>Fe` hết, `H_2 SO_4` dư, tính theo `Fe`
\(n_{H_2SO_4\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=n\cdot M=0,1\cdot98=9,8\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\)
nFe3O4 = \(\dfrac{46,4}{232}\)= 0,2mol
nH2 = \(\dfrac{8,96}{22,4}\) = 0,4 mol
Fe3O4 + 4H2 ->4 H2O + 3Fe
0,2(dư);0,4(hết)
=>mFe3O4(dư) = 0,1 . 232 = 23,2 g