Đốt cháy hoàn toàn 3,1g P trong bình chứa O2, sản phẩm tạo thành là P2O5.Tính thể tích O2 cần dùng cho phản ánh ứng trên (đktc)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{19,5}{65}=0,3mol\)
\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)
0,3 0,15 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -t°-> 2P2O5
0,1---> 0,125--->0,05
VO2 = 0,125 . 22,4 = 2,8 (l)
mP2O5 = 0,05 . 142 = 7,1 (g)
\(n_P=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,1 0,125 0,05
\(V_{O_2}=0,125\cdot22,4=2,8l\)
\(m_{P_2O_5}=0,05\cdot142=7,1g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{19,5}{65}=0,3mol\)
\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)
0,3 0,15 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)
b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
0,1 0,15 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,1.122,5=12,25g\)
a) PTHH: 2Zn + O2 → 2ZnO
2 1 2
0,3 0,15 0,3
nZn = \(\dfrac{m}{M}\) = \(\dfrac{19,5}{65}\) = 0,3 (mol)
mO2 = n.M = 0,15 . 16 = 2,4 (g)
VO2 = m . 22,4 = 2,4 . 22,4 = 53,76 (l)
b) 2KClO3 → 2KCl + 3O2 ↑
0,1 0,1 0,15
mKClO3 = n . M = 0,1 . 122,5 = 12,25 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
3Fe+2O2-to>Fe3O4
0,225--0,15
n Fe=\(\dfrac{12,6}{56}\)=0,225 mol
VO2=0,15.22,4=3,36l
2KClO3-to>2KCl+3O2
0,1---------------------0,15
=>m KClO3=0,1.122,5=12,25g
\(a,3Fe+2O_2\rightarrow Fe_3O_4\)
\(b,\)
Ta có : \(n_{Fe}=\dfrac{m}{M}=\dfrac{126}{56}=2,25\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.2,25=1,5\left(mol\right)\)
\(\Rightarrow VO_2=33,6\left(l\right)\)
\(c,\)
\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)
Theo \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\\ Theo.PTHH:n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ V_{O_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
\(b,Theo.PTHH:n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,2.0,1\left(mol\right)\\ m_{P_2O_5}=n.M=0,1.142=14,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
_____0,4____0,5_____0,2 (mol)
a, \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
b, \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(0.1........0.125\)
\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)