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nP= 7,44/31=0,24(mol)
nO2=6,16/22,4=0,275(mol)
PTHH:4 P + 5 O2 -to->2 P2O5
Ta có: 0,24/4 > 0,275/5
=> O2 hết, P dư, tính theo nO2
nP(p.ứ)= 0,275 x 4/5= 0,22(mol)
=>nP(dư)=0,24-0,22=0,02(mol)
=>mP(dư)=0,02.31= 0,62(g)
nP2O5= 2/5 x 0,275= 0,11(mol)
=> mP2O5= 142 x 0,11= 15,62(g)
\(n_P=\dfrac{7,44}{31}=0,24\left(mol\right)\)
\(n_{O_2}=\dfrac{6,16}{22,4}=0,275\left(mol\right)\)
PTHH : \(4P+5O_2\rightarrow2P_2O_5\)
Ban đầu : 0,24 0,275 (mol)
Phản ứng : 0,22 0,275 0,11 (mol)
Sau phản ứng : 0,02 0 0,11 (mol)
\(m_P=0,02.31=0,62\left(g\right)\)
\(m_{P_2O_5}=0,11.142=15,62\left(g\right)\)
\(n_P=\dfrac{6,2}{31}=0,2mol\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
Xét: \(\dfrac{0,2}{4}\) < \(\dfrac{0,3}{5}\) ( mol )
0,2 0,1 ( mol )
\(m_{P_2O_5}=0,1.142=14,2g\)
`PTHH: 4P + 5O_2` $\xrightarrow[]{t^o}$ `2P_2 O_5`
`n_P = [ 6,2 ] / 31 = 0,2 (mol)`
`n_[O_2] = [ 6,72 ] / [ 22,4 ] = 0,3 (mol)`
Ta có: `[ 0,2 ] / 4 < [ 0,3 ] / 5`
`->P` hết ; `O_2` dư
Theo `PTHH` có: `n_[P_2 O_5] = 1 / 2 n_P = 1 / 2 . 0,2 = 0,1 (mol)`
`-> m_[P_2 O_5] = 0,1 . 142 = 14,2 (g)`
a) \(PTHH:4P+5O_2\) → \(2P_2O_5\)
\(n_{O_2}=\dfrac{V_{O_2\left(đktc\right)}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PTHH:
⇒ \(n_P=\dfrac{4}{5}n_{O_2}=\dfrac{4}{5}.0,5=0,4\left(mol\right)\)
\(m_P=n.M=0,4.31=12,4\left(g\right)\)
b) Theo PTHH:
⇒ \(n_{P_2O_5}=\dfrac{1}{2}.n_p=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)
\(m_{P_2O_5}=n.M=0,2.142=28,4\left(g\right)\)
nP = 6.2/31 = 0.2 (mol)
nO2 = 6.72/22.4 = 0.3 (mol)
4P + 5O2 -to-> 2P2O5
0.2___0.25_____0.1
mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g)
mP2O5 = 0.1*142 = 14.2 (g)
Ta có: \(n_P=\dfrac{6.2}{31}=0.29mol\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3mol\)
PTHH:
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
ta có:
\(\left\{{}\begin{matrix}\dfrac{n_{P\left(bra\right)}}{nP_{\left(pthh\right)}}=\dfrac{0.2}{4}=0.05\\\dfrac{n_{O_2\left(bra\right)}}{n_{O_2}\left(pthh\right)}=\dfrac{0.3}{5}=0.06\end{matrix}\right.\)
=> \(O_2\) dư
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 ----------->2
0.2---------->0.1=nP2O5
=>\(m_{P_2O_5}=142.0.1=14.2\left(g\right)\)
a) Theo PTHH:
nO2=54nP=54.0,2=0,25nO2=54nP=54.0,2=0,25 (mol)
Thể tích khí oxi tham gia phản ứng (đktc) là:
VO2=0,25.22,4=5,6VO2=0,25.22,4=5,6 (l)
b)
nP=6,231=0,2nP=6,231=0,2 (mol)
Theo PTHH:
nP2O5=12nP=12.0,2=0,1nP2O5=12nP=12.0,2=0,1 (mol)
Khối lượng P2O5P2O5 thu được sau phản ứng là:
mP2O5=0,1.142=14,2mP2O5=0,1.142=14,2 (g)
a) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
_____0,2-->0,25------>0,1
=> VO2 = 0,25.22,4 = 5,6 (l)
b) mP2O5 = 0,1.142 = 14,2 (g)
\(n_P=\frac{7,44}{31}=0,24\left(mol\right)\)
\(n_{O2}=\frac{6,6}{22,4}=\frac{33}{112}\left(mol\right)\)
a. \(PTHH:4P+5O_2\rightarrow2P_2O_5\)
Trước_____ 0,24___33/112
Phứng___33/140__33/112
Sau____0,00428__ 0 _____33/280
Nên P dư, \(m_R=0,00428.31=0,13268\left(g\right)\)
b. \(m_{P2O5}=\frac{142.33}{280}=16,74\left(g\right)\)
a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)
0,4-->0,5----->0,2
b) \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
\(n_P=\dfrac{5}{31}=0,16mol\)
\(V_{O_2}=\dfrac{V_{kk}}{5}=\dfrac{2,8}{5}=0,56l\)
\(n_{O_2}=\dfrac{0,56}{22,4}=0,025mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
\(\dfrac{0,16}{4}\)< \(\dfrac{0,25}{5}\) ( mol )
0,16 0,08 ( mol )
\(m_{P_2O_5}=0,08.142=11,36g\)
nP = 7,44/31 = 0,24 (mol)
nO2 = 6,16/22,4 = 0,275 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,24/4 > 0,275/5 => P dư
nP2O5 = 0,275 : 5 . 2 = 0,11 (mol)
mP2O5 = 0,11 . 142 = 15,62 (g)
\(4P+5O_2->2P_2O_5\)
4 5 2
0,24 0,3 0,12 (mol)
\(n_P=\dfrac{m}{M}=\dfrac{7,44}{31}=0,24\left(mol\right)\)
\(m_{P_2O_5}=n\text{×}M=0,12\text{×}\left(31\text{×}2+16\text{×}5\right)=0,12\text{×}142=17,04\left(g\right)\)