Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Fe2O3+3H2-to>2Fe+3H2O
0,1125---0,3375----0,225 mol
n Fe=0,225 mol
=>m Fe2O3=0,1125.160=18g
=>VH2=0,3375.22,4=7,56l
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: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
Phương trình hóa học của phản ứng:
Fe2O3 + 3H2 → 2Fe + 3H2O.
Khử 1 mol Fe2O3 cho 2 mol Fe.
x mol Fe2O3 → 0,2 mol.
x = 0,2 : 2 =0,1 mol.
m = 0,1 .160 =16g.
Khử 1 mol Fe2O3 cần 3 mol H2.
Vậy khử 0,1 mol Fe2O3 cần 0,3 mol H2.
V= 0,3 .22.4 = 6,72l.
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, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)
\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)
nFe=0,2(mol)
a) PTHH: Fe2O3 + 3 H2 -to-> 2 Fe + 3 H2O
0,1_____________0,3____0,2(mol)
b) mFe2O3=160.0,1=16(g)
c) V(H2,đktc)=0,3.22,4=6,72(l)
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
\(b.m_{Fe_2O_3}=n.M=0,43.\left(56.2+16.3\right)=68,8\left(g\right)\\ c.V_{H_2}=n.24,79=0,645.24,79=15,98955\left(l\right).\)
a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,2\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)
c, \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)
d, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
a)
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
b) $n_{Fe} = \dfrac{22,4}{56} = 0,4(mol)$
Theo PTHH : $n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe} = 0,2(mol)$
$m_{Fe_2O_3} = 0,2.160 = 32(gam)$
c) $n_{H_2} = \dfrac{3}{2}n_{Fe} = 0,6(mol)$
$V_{H_2} = 0,6.22,4 = 13,44(lít)$
d) $2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
$V_{O_2} = \dfrac{1}{2}V_{H_2} = 6,72(lít)$
$V_{kk} = 6,72 : 20\% = 33,6(lít)$
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
1 3 2 3 ( mol )
0,1 0,3 0,2 ( mol )
\(m_{Fe_2O_3}=n_{Fe_2O_3}.M_{Fe_2O_3}=0,1.160=16g\)
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)