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\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\
pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
2,25 1,5
=> \(V_{O_2}=1,5.22,4=33,6\left(L\right)\)
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
1 1,5
=> \(m_{KClO3}=122,5\left(g\right)\)
nFe = 33,6 : 56 = 0,6 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,6--> 0,4------->0,2 (mol)
=> vO2 = 0,4.22,4 = 8,96 (mol)
=> mFe3O4 = 0,2.232 = 46,4 (g)
pthh : 2KClO3 -t--> 2KClO3 + 3O2
0,267<-----------------------0,4(mol)
mKClO3= 0,267 .122,5 = 32,67 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
\(a.PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2 1
0,9 0,6 0,3
\(b.V_{O_2}=n.24,79=0,6.24,79=14,874\left(l\right)\)
\(c.m_{Fe_3O_4}=n.M=0,3.\left(56.3+16.4\right)=69,6\left(g\right)\)
\(d.V_{O_2}=14,874\left(l\right)\\ \Rightarrow n_{O_2}=\dfrac{V}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2 2 3
0,6 0,6 0,9
\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)
\(n_P=\dfrac{7,44}{31}=0,24mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,24 0,3 0,12
\(V_{O_2}=0,3\cdot22,4=6,72l\)
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
0,2 0,3
\(m_{KClO_3}=0,2\cdot122,5=24,5g\)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
Theo PT: \(n_P=2n_{P_2O_5}=0,1\left(mol\right)\)
\(\Rightarrow m_P=0,1.31=3,1\left(g\right)\)
\(n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
c, Có: \(V_{O_2\left(dư\right)}=2,8.15\%=0,42\left(l\right)\)
\(\Rightarrow V_{O_2}=2,8+0,42=3,22\left(l\right)\)
PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a) Ta có: \(\left\{{}\begin{matrix}n_{P_2O_5}=\dfrac{10,65}{142}=0,075\left(mol\right)\\\Sigma n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_P=0,15mol\\n_{O_2\left(dư\right)}=0,0625mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_P=0,15\cdot31=4,65\left(g\right)\\m_{O_2\left(dư\right)}=0,0625\cdot32=2\left(g\right)\end{matrix}\right.\)
b) Ta có: \(n_{O_2\left(pư\right)}=0,1875mol\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(pư\right)}=0,1875\cdot32=6\left(g\right)\\V_{O_2\left(pư\right)}=0,1875\cdot22,4=4,2\left(l\right)\end{matrix}\right.\)