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\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(0.1.......0.125.....0.05\)
\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(m_{P_2O_5}=0.05\cdot142=7.1\left(g\right)\)
nP= 3,1 / 31 =0,1 mol
2P + 5/2O2 → P2O5
0,1 0,125 0,05 mol
VO2=0,125.22,4=2,8 l
b) mP2O5=0,05.142=7,1 g
a. \(n_P=\dfrac{3.1}{31}=0,1\left(mol\right)\)
PTHH : 4P + 5O2 ---to----> 2P2O5
0,1 0,125 0,05
b. \(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
c. \(V_{O_2}=0,125.22,4=2,8\left(l\right)\\ \Rightarrow V_{kk}=2,8.5=14\left(l\right)\)
Bài 1:
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(4P+5O_2\rightarrow2P_2O_5\)
0,24.... 0,3 .... 0,12 (mol)
\(m_P=0,24.31=7,44\left(g\right)\)
\(m_{P_2O_5}=0,12.142=17,04\left(g\right)\)
Bài 2:
\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
0,8 .... 0,6 ...... 0,4 (mol)
\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
2KMnO4-to>K2MnO4+MnO2+O2
0,3-----------------0,15-----0,15------0,15 mol
n KMnO4=\(\dfrac{47,4}{158}\)=0,3 mol
=>mcr=0,15.197.0,15.87=42,6g
=>VO2=0,15.22,4=3,36l
b) 4P+5O2-to>2P2O5
0,1--------------0,05
nP=\(\dfrac{3,1}{31}\)=0,1 mol
->O2 dư
=>m P2O5=0,05.142=7,1g
mKMnO4 = 47,4/158 = 0,3 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15
m = 0,15 . 197 + 0,15 . 87 = 85,2 (g)
V = VO2 = 0,15 . 22,4 = 3,36 (l)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,1/4 < 0,15/5 => O2 dư
nP2O5 = 0,1/2 = 0,05 (mol)
mP2O5 = 0,05 . 142 = 7,1 (g)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
_____0,1-->0,125---->0,05
=> mP2O5 = 0,05.142 = 7,1 (g)
c) VO2 = 0,125.22,4 = 2,8(l)
\(n_P=\dfrac{m}{M}=\dfrac{31}{31}=1\left(mol\right)\\ n_{O_2\left(dktc\right)}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
ti lệ: 4 : 5 : 2
n(mol) 1 0,5
n(mol p/ư): 0,4<--0,5------>0,2
\(\dfrac{n_P}{4}>\dfrac{n_{O_2}}{5}\left(\dfrac{1}{4}>\dfrac{0,5}{5}\right)\)
`=>` `O_2` hết, `P` dư, tính theo`O_2`
\(n_{P\left(dư\right)}=1-0,4=0,6\left(mol\right)\)
\(m_{P\left(dư\right)}=n\cdot M=0,6\cdot31=18,6\left(g\right)\\ m_{P_2O_5}=n\cdot M=0,2\cdot\left(31\cdot2+16\cdot5\right)=28,4\left(g\right)\)
\(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
0,2 0,25 0,1 ( mol )
a, \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
b, \(m_P=0,2.31=6,2\left(g\right)\)
c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{1}{6}\) 0,25 ( mol )
\(m_{KClO_3}=\dfrac{1}{6}.122,5=\dfrac{245}{12}\approx20,42\left(g\right)\)
\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\b,n_{P_2O_5}=\dfrac{2}{5}.0,25=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\\c,V_{kk\left(đktc\right)}=4.5,6=28\left(lít\right) \)
a) nP=0,1(mol)
PTHH: 4P + 5 O2 -to-> 2 P2O5
nO2= 5/4 . nP=5/4. 0,1= 0,125(mol)
=> V(O2,dktc)=0,125.22,4=2,8(l)
b) nP2O5= nP/2= 0,1/2= 0,05(mol)
=>mP2O5=142.0,05=7,1(g)