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4P + 5O2 ----> 2P2O5
0,24 -> 0,3 ---> 0,12 (mol)
nP = \(\dfrac{7,44}{31}\)= 0,24 (mol)
VH2 = 0,3 . 22,4 = 6,72 (l)
2KClO3 ---> 2KCl + 3O2
0,2 <------------- 0,3 (mol)
mKClO3 = 0,2 . (39 + 35,5 + 16.3)
= 24,5 (g)
Vui lòng kiểm tra lại kết quả dùm, thank you.
nP = 7,44 : 31 = 0,24 ( mol)
pthh : 4P + 5O2 -t--> 2P2O5
0,24->0,3 (mol)
=> VO2 =0,3 . 22,4 = 6,72 (l)
pthh : 2KClO3 -t--> 2KCl + 3O2
0,2<-------------------0,3 (mol)
=> mKClO3 = 0,2 .122,5 = 24,5 (g)
a. \(n_P=\frac{6,2}{31}=0,2mol\)
\(V_{O_2}=V_{kk}.\frac{1}{5}=\frac{18,48}{5}=3,696l\)
\(n_{O_2}=\frac{3,696}{22,4}=0,165mol\)
PTHH: \(4P+5O_2\xrightarrow{t^o}2P_2O_5\)
Tỷ lệ \(\frac{0,2}{4}>\frac{0,165}{5}\)
Vậy P dư
\(n_{P\left(\text{phản ứng }\right)}=\frac{4}{5}n_{O_2}=0,132mol\)
\(n_{P\left(dư\right)}=0,2-0,132=0,068mol\)
\(\rightarrow m_{P\left(dư\right)}=0,068.31=2,108g\)
b. \(n_{P_2O_5}=\frac{2}{5}n_{O_2}=0,066mol\)
\(\rightarrow m_{P_2O_5}=0,066.142=9,372g\)
c. PTHH: \(2KClO_3\xrightarrow{t^o}2KCl+3O_2\)
\(n_{KClO_3}=\frac{2}{3}n_{O_2}=0,11mol\)
\(\rightarrow m_{KClO_3}=0,11.122,5=13,475g\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)
b, \(V_{kk}=5V_{O_2}=28\left(l\right)\)
c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
a.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,3 0,6 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,6.22,4=13,44l\)
b.
\(n_P=\dfrac{m_P}{M_P}=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,1 0,05 ( mol )
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1g\)
\(PTHH:4P+5O_2->2P_2O_5\)
Số mol của Photpho: \(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(PTHH:4P+5O_2->2P_2O_5\)
4 mol 5 mol 2 mol
0,2 mol --------> 0,1 mol
Khối lượng Diphotpho Pentaoxit: \(\left(M_{P_2O_5}=142g/mol\right)\)
\(m_{P_2O_5}=n.M=0,1.142=14,2\left(g\right)\)
b) \(PTHH:4P+5O_2->2P_2O_5\)
4 mol 5 mol
0,2 mol -> 0,25 mol
Thể tích khí oxi (đktc) cần dùng: \(V_{O_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
Chúc bn học tốt nha ^^
\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)
Ta có:
nP= \(\frac{m_P}{M_P}=\frac{12,4}{31}=0,4\left(mol\right)\)
PTHH:4 P + 5O2 -> 2P2O5
a) Theo PTHH và đề bài, ta có:
\(n=\frac{5.n_P}{4}=\frac{5.0,4}{4}=0,5\left(mol\right)\)
=> \(V_{O_2\left(đktc\right)}=n_{O_2}.22,4=0,5.22,4=11,2\left(l\right)\)
b) Ta có:
\(n_{P_2O_5}=\frac{2.n_P}{4}=\frac{2.0,4}{4}=0,2\left(mol\right)\)
=> \(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,2.142=28,4\left(g\right)\)
a) PTHH: 4P + 5O2 =(nhiệt)=> 2P2O5
nP = 12,4 / 31 = 0,4 mol
=> nO2 = 0,5 (mol)
=> VO2(đktc) = 0,5 x 22,4 = 11,2 lít
b) nP2O5 = \(\frac{1}{2}n_P=0,2\left(mol\right)\)
=> VP2O5(đktc) = 0,2 x 22,4 = 4,48 lít
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)
\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)
\(0.1.......0.125.....0.05\)
\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(m_{P_2O_5}=0.05\cdot142=7.1\left(g\right)\)
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
\(n_{P_2O_5}=\dfrac{28.4}{142}=0.2\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(.......0.5......0.2\)
\(V_{O_2}=0.5\cdot22.4=11.2\left(l\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(\dfrac{1}{3}...................0.5\)
\(m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)
a) nP2O5= 28,4/142= 0,2(mol)
PTHH: 4P + 5 O2 -to-> 2 P2O5
nO2= 5/2 . 0,2= 0,5(mol)
=>V(O2,đktc)=0,5.22,4= 11,2(l)
b) 2KClO3 -to-> 2 KCl + 3 O2
1/3_________________0,5(mol)
mKClO3= 1/3 . 122,5\(\approx\) 40,833(g)