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a) 4P + 5O2 --to--> 2P2O5
b) \(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,8--------------->0,4
=> mP2O5 = 0,4.142 = 56,8 (g)
c)
PTHH: P2O5 + 3H2O --> 2H3PO4
0,4--------------->0,8
=> mH3PO4 = 0,8.98 = 78,4 (g)
\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\b,n_{P_2O_5}=\dfrac{2}{5}.0,25=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\\c,V_{kk\left(đktc\right)}=4.5,6=28\left(lít\right) \)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
\(n_{P_2O_5}=\dfrac{28,4}{142}=0,2mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,5 0,2 ( mol )
\(m_{O_2}=0,5.32=16g\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
0,2 0,4 ( mol )
\(m_{H_3PO_4}=0,4.98=39,2g\)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
d, Vì: VO2 = 1/5Vkk
\(\Rightarrow V_{kk}=5V_{O_2}=14\left(l\right)\)
Bạn tham khảo nhé!
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) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
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)
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)
\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
4P+5O2-to>2P2O5
0,4--0,5-----------0,2 mol
P2O5+3H2O->2H3PO4
0,2------------------0,4
n O2=\(\dfrac{11,2}{22,4}\)=0,5 mol
=>m P=0,4.31=12,4g
=>m H3PO4=0,4.98=39,2g