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a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)
c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)
c, \(n_{H_2}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{15}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\)
d, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)
\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)
a + b)
\(n_{Fe_3O_2}=\dfrac{m_{Fe_3O_2}}{M_{Fe_3O_2}}=\dfrac{9,6}{200}=0,048\left(mol\right)\)
Gọi kim loại thu được là A
PTHH: \(Fe_3O_2+H_2\rightarrow A+H_2O\)
Theo PT: 1mol __1mol__1mol_1 mol
Theo đề: 0,048 mol_0,048 mol_0,048 mol_0,048 mol
\(n_{H_2}=\dfrac{n_{Fe_3O_2}.1}{1}=0,048\left(mol\right)\)
\(V_{H_2}=n_{H_2}.22,4=0,048.22,4=1,0752\left(l\right)\)
`a)`
`3H_2+Fe_2 O_3` $\xrightarrow{t^o}$ `2Fe + 3H_2 O`
`0,75` `0,25` `0,5` `0,75` `(mol)`
`n_[Fe_2 O_3]=40/160=0,25(mol)`
`b)V_[H_2]=0,75.22,4=16,8(l)`
`c)m_[Fe]=0,5.56=28(g)`
`d)V_[H_2 O]=0,75.22,4=16,8(l)`
a, `3H_2 + Fe_2O_3 -> 2Fe + 3H_2O`.
`=> n_(Fe_2O_3) = (m(Fe_2O_3))/(M_(Fe_2O_3)) = 40/160 = 0,25 mol`.
b,` n_(H_2) = 0,25 xx 3 = 0,75 mol`.
`V_(H_2) = 0,75 xx 22,4 = 16,8l`.
c, `n_(Fe) = 0,25 xx 3 = 0,5 mol`.
`m_(Fe) = n_(Fe) . M_(Fe) = 0,5 xx 56 = 28 g`.
d, `n_(H_2O) = 0,25 xx 3 = 0,75 mol`.
`V_(H_2O) = 0,75 xx 22,4 = 16,8 l`.
a) 3H2+ Fe2O3→ 3H2O+ 2Fe
(mol) 0,9 0,3 0,6
b) nFe=\(\dfrac{m}{M}=\dfrac{33,6}{56}=0,6\left(mol\right)\)
→ \(V_{H_2}=n.22,4=0,9.22,4=20,16\left(lít\right)\)
\(m_{Fe_2O_3}=n.M=0,3.160=48\left(g\right)\)
c) 3Fe+ 2O2→ Fe3O4
(mol) 0,3 0,2 \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Xét tỉ lệ:
Fe O2
\(\dfrac{0,6}{3}\) > \(\dfrac{0,2}{2}\)
→ Fe dư, O2 phản ứng hết.
→nFe(còn lại)=nFe(ban đầu)-nFe(phản ứng)=0,6-0,3=0,3
=> mFe(còn lại)=n.M=0,3.56=16,8(g)
Vậy sau khi phản ứng Fe dư và dư 16,8g.
a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe}=\dfrac{24}{56}=\dfrac{3}{7}\left(mol\right)\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{14}\left(mol\right)\Rightarrow m_{Fe_2O_3}=\dfrac{3}{14}.160=\dfrac{240}{7}\left(g\right)\)
c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=\dfrac{9}{14}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{9}{14}.22,4=14,4\left(l\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{24}{56}\approx0,43\left(mol\right)\\ a.PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
2 3 2 3
0,43 0,645 0,45 0,645
\(b.m_{Fe_2O_3}=n.M=0,43.\left(56.2+16.3\right)=68,8\left(g\right)\\ c.V_{H_2}=n.24,79=0,645.24,79=15,98955\left(l\right).\)
\(n_{Fe_2O_3=}=\dfrac{24}{160}=0,15mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,15 0,45 0,3 0,45
\(V_{H_2}=0,45\cdot224,=10,08l\)
\(m_{Fe}=0,3\cdot56=16,8g\)