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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) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH :
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,2 0,15 0,1
\(b,m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
\(c,V_{O_2}=0,15.24,79=3,7185\left(l\right)\)
a) PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)
b) \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{13,5}{27}=0,5\left(mol\right)\)
Theo PTHH: \(n_{Al_2O_3}=\dfrac{0,5.2}{4}=0,25\left(mol\right)\)
Khối lượng sản phẩm tạo thành: \(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,25.102=25,5\left(g\right)\)
c) Theo PTHH: \(n_{O_2}=\dfrac{0,5.3}{4}=0,375\left(mol\right)\)
Thể tích không khí cần dùng: \(V_{O_2}=n_{O_2}.22,4=0,375.22,4=8,4\left(l\right)\)
4P+5O2-to>2P2O5
0,04----0,05----0,02
n P=0,04 mol
=>m P2O5=0,02.142=2,84g
=>VO2=0,05.22,4=1,12l
c)
2KMnO4-to>K2MnO4+MnO2+O2
0,1----------------------------------------0,05
H=10%
m KMnO4=0,1.158.110%=17,28g
\(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{T^O}2P_2O_5\)
0,04 0,05 0,02
=> \(\left\{{}\begin{matrix}m_{P_2O_5}=0,02.142=2,84\left(g\right)\\V_{O_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)
\(pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
=> \(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(m_{KMnO_4\left(d\text{ùng}\right)}=15,8.110\%=17,38\left(g\right)\)
a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
a, Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,075\left(mol\right)\Rightarrow V_{O_2}=0,075.22,4=1,68\left(l\right)\)
c, Có lẽ đề cho 0,112 chứ không phải 0,1121 bạn nhỉ?
Ta có: \(n_{O_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{4}>\dfrac{0,005}{3}\), ta được Al dư.
Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{300}\left(mol\right)\Rightarrow m_{Al_2O_3}=\dfrac{1}{300}.102=0,34\left(g\right)\)
a) \(3Fe+2O_2-t^o->Fe_3O_4\)
b) \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)
=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)
c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)
=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)
=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,6 0,3 0,6 ( mol )
\(m_{H_2O}=0,6.18=10,8g\)
\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=33,6l\)
a. \(n_{CH_4}=\dfrac{0.56}{22,4}=0,025\left(mol\right)\)
PTHH : CH4 + 2O2 -> CO2 + 2H2O
0,025 0,05 0,025
b. \(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
c. \(V_{CO_2}=0,025.22,4=0,56\left(l\right)\)