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a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2-->0,25------->0,1
=> mP2O5 = 0,1.142 = 14,2(g)
c) VO2 = 0,25.22,4 = 5,6(l)
$\rm n_P=\dfrac{6,2}{31}=0,2(mol)$
$\rm a)PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5$
$\rm b)$ Theo PT: $\rm n_{O_2}=1,25n_P=0,25(mol)$
$\rm V_{O_2}=0,25.22,4=5,6(lít)$
a, Số mol P là
n = m/M = 6,2/31 = 0,2 (mol)
a, PTHH : 4P + 5O2 -t0> 2P2O5
4 5 2
0,2 mol -> 0,25 mol 0,1 mol
b, Thể tích O2 theo đktc là :
V = n . 22,4 = 0,25 . 22,4 = 5,6 (l)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH : 4P + 5O2 -> 2P2O5
=> \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
=> \(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Theo ĐLBTKL
\(m_P+m_{O_2}=m_{P_2O_5}\\ =>m_{O_2}=14,2-6,2=8\left(g\right)\)
=> \(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\\ V_{O_2}=0,25.22,4=5,6\left(l\right)\)
\(n_P=\dfrac{m}{M}=\dfrac{31}{31}=1\left(mol\right)\\ n_{O_2\left(dktc\right)}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
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`
\(n_{P\left(dư\right)}=1-0,4=0,6\left(mol\right)\)
\(m_{P\left(dư\right)}=n\cdot M=0,6\cdot31=18,6\left(g\right)\\ m_{P_2O_5}=n\cdot M=0,2\cdot\left(31\cdot2+16\cdot5\right)=28,4\left(g\right)\)
PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{9,3}{31}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,375\left(mol\right)\\n_{P_2O_5}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,375\cdot22,4=8,4\left(g\right)\\m_{P_2O_5}=0,15\cdot142=21,3\left(g\right)\end{matrix}\right.\)
\(n_{O_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ 0,12........0,15.........0,06\left(mol\right)\\ m_P=0,12.31=3,72\left(g\right)\)
a, \(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_{O_2}=\dfrac{5}{4}n_P=1\left(mol\right)\) \(\Rightarrow V_{O_2}=1.22,4=2,24\left(l\right)\)
b, \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,4\left(mol\right)\Rightarrow m_{P_2O_5}=0,4.142=56,8\left(g\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=2\left(mol\right)\Rightarrow m_{KMnO_4}=2.158=316\left(g\right)\)
\(n_P=\dfrac{m}{M}=\dfrac{24,8}{31}=0,8\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 5 2
0,8 1 0,4
\(a.V_{O_2}=n.24,79=1.24,79=24,79\left(l\right)\\ b.m_{P_2O_5}=n.M=0,4.\left(31.2+16.5\right)=56,8\left(g\right)\)
\(c.PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2\downarrow+O_2\uparrow\)
2 1 1 1
0,8 0,4 0,4 0,4
\(m_{KMnO_4}=n.M=0,8.\left(39+55+16.4\right)=126,4\left(g\right).\)
\(a.n_P=\dfrac{6,2}{31}=0,2mol\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\)
\(n_{P_2O_5}=0,2.2:4=0,1mol\\ m_{P_2O_5}=0,1.142=14,2g\)
\(b.n_{P_2O_5}=\dfrac{35,5}{142}=0,25mol\\ n_P=0,25.2=0,5mol\\ m_P=0,5.31=15,5g\\ n_{O_2}=\dfrac{0,25.5}{2}=0,625mol\\ V_{O_2}=0,625.24,79=15,49375l\)