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a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)
PTHH : 3Fe + 2O2 -to-> Fe3O4
0,09 0,06 0,03
\(m_{Fe}=0,09.56=5,04\left(g\right)\)
\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)
b. PTHH : 2KCl + 3O2 -> 2KClO3
0,06 0,04
\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)
4K + O2 --to--➢ 2K2O (1)
\(n_{K_2O}=\dfrac{16,8}{94}=\dfrac{42}{235}\left(mol\right)\)
Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{K_2O}=\dfrac{1}{2}\times\dfrac{42}{235}=\dfrac{21}{235}\left(mol\right)\)
2KClO3 --to--➢ 2KCl + 3O2 (2)
\(n_{O_2\left(1\right)}=n_{O_2\left(2\right)}\)
Theo PT2: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\times\dfrac{21}{235}=\dfrac{14}{235}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{14}{235}\times122,5=7,3\left(g\right)\)
\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(a.\)
\(2H_2+O_2\underrightarrow{^{^{t^0}}}2H_2O\)
\(0.5.......0.25\)
\(m_{O_2}=0.25\cdot32=8\left(g\right)\)
\(b.\)
\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)
\(0.5............................................0.25\)
\(m_{KMnO_4}=0.5\cdot158=79\left(g\right)\)
mMg = 3.6/24 = 0.15 (mol)
2Mg + O2 -to-> 2MgO
0.15__0.075____0.15
mMgO= 0.15*40 = 6 (g)
VO2 = 0.075*22.4 = 1.68 (l)
2KClO3 -to-> 2KCl + 3O2
0.05_______________0.075
mKClO3 = 0.05*122.5 = 6.125 (g)
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)
4P (0,2 mol) + 5O2 (0,25 mol) \(\underrightarrow{t^o}\) 2P2O5.
a) Thể tích không khí cần dùng ở đktc là 0,25.22,4.5=28 (lít).
b) 2KMnO4 (0,5 mol) \(\underrightarrow{t^o}\) K2MnO4 + MnO2 + O2 (0,25 mol).
Khối lượng cần tìm là 0,5.158=79 (g).
a.\(n_P=\dfrac{6,2}{31}=0,2mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,2 0,25 ( mol )
\(V_{kk}=V_{O_2}.5=\left(0,25.22,4\right).5=28l\)
b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,5 0,25 ( mol )
\(m_{KMnO_4}=0,5.158=79g\)
nK=0,2(mol)
PTHH: 4K + O2 -to-> 2 K2O
nK2O= 0,1(mol) => mK2O=0,1.94=9,4(g)
nO2=0,05(mol) -> V(O2,đktc)=0,05.22,4=1,12(l)
V(kk,dktc)=5.V(O2,dktc)=5.1,12=5,6(l)
a)
n K2O = 16,8/94 = 0,18 (mol)
PTHH: 4K + O2 ---> 2K2O (1)
0,36 0,09 0,18 (mol)
Theo PTHH(1), có:
n K=2nK2O =0,18.2 =0,36(mol)
=> mK=0,36.39=14,04(g)
b)
PTHH: 2KClO3 ---> 2KCl + 3O2 (2)
Theo Pthh(1) và (2) ,có:
nO2(2)=nO2(1)=0,09 (mol)
=> nKClO3 =2/3nO2(2)=2/3.0,09=0,06 (mol)
=> m KClO3=122,5. 0,06=7,35(g)
4K + O2 --to--➢ 2K2O (1)
a) \(n_{K_2O}=\dfrac{16,8}{94}=\dfrac{42}{235}\left(mol\right)\)
Theo PT1: \(n_K=2n_{K_2O}=2\times\dfrac{42}{235}=\dfrac{84}{235}\left(mol\right)\)
\(\Rightarrow m_K=\dfrac{84}{235}\times39=13,94\left(g\right)\)
b) 2KClO3 --to--➢ 2KCl + 3O2 (2)
Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{K_2O}=\dfrac{1}{2}\times\dfrac{42}{235}=\dfrac{21}{235}\left(mol\right)=n_{O_2\left(2\right)}\)
Theo PT2: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\times\dfrac{21}{235}=\dfrac{14}{235}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{14}{235}\times122,5=7,3\left(g\right)\)