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a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=11,2\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
a.b.c.\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}\dfrac{4,8}{24}=0,2mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
0,2 0,1 0,2 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24l\)
\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)
d. Sửa đề: tính khối lượng KMnO4
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,2 0,1 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,2.158=31,6g\)
a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
0,3--->0,2----->0,1
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)
c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)
\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,44<------------------------------------0,22
\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)
Theo gt ta có: $n_{H_2}=0,75(mol)$
a, $2H_2+O_2\rightarrow 2H_2O$
Ta có: $n_{O_2}=0,5.n_{H_2}=0,375(mol)\Rightarrow V_{O_2}=8,4(l)\Rightarrow V_{kk}=42(l)$
b, $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$
Ta có: $n_{KMnO_4}=2.n_{O_2}=0,75(mol)\Rightarrow m_{KMnO_4}=118,5(g)$
a)
\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{V_{H_2}}{2} = \dfrac{16,8}{2} = 8,4(lít)\\ V_{không\ khí} = \dfrac{8,4}{20\%} = 42(lít)\)
b)
\(n_{O_2} = \dfrac{8,4}{22,4} = 0,375(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,75(mol)\\ \Rightarrow m_{KMnO_4} = 0,75.158 = 118,5(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,25(mol)\\ \Rightarrow m_{KClO_3} = 0,25.122,5 = 30,625(gam)\)
a.b.\(n_{Mg}=\dfrac{m}{M}=\dfrac{6,4}{24}=\dfrac{4}{15}mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
4/15 2/15 ( mol )
\(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}l\)
c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
4/15 2/15 ( mol )
\(m_{KMnO_4}=n.M=\dfrac{4}{15}.158=\dfrac{632}{15}g\)
nMg = 6,4 : 24= 0,26(mol)
pthh : 2Mg+O2 -t--> 2MgO
0,26 --> 0,13 (mol )
=> VO2(đktc) = 0,13.22,4=2,912(l)
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2
0,26<------------------------------0,13(mol)
=> mKMnO4 = 0,26.158= 41,08(g)
a) nFe = 16,8/56 = 0,3 (mol)
PTHH: 3Fe + 2O2 -> (t°) Fe3O4
Mol: 0,3 ---> 0,2 ---> 0,1
mFe3O4 = 0,1 . 232 = 23,2 (g)
b) VO2 = 0,2 . 22,4 = 4,48 (l)
Vkk = 4,48 . 5 = 22,4 (l)
c) H = 100% - 20% = 80%
nO2 (LT) = 0,2 : 80% = 0,25 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
nKMnO4 = 0,25 . 2 = 0,5 (mol)
mKMnO4 = 0,5 . 158 = 79 (g)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)
PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)
Ta có: \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=0,4\left(mol\right)\\n_{O_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,4\cdot22,4=8,96\left(l\right)\\V_{O_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)
a, pthh:C2H5OH + 7/2O2-> 2CO2 + 3H2O(nhiệt độ cao)
b, ta có nC2H5OH=9,2/46=0,2 mol
từ PTHH:
C2H5OH+7/2O2->2CO2+3H2O
1mol 2mol
0,2mol x mol
=> 1/0,2=2/x<=>x=nCO2=0,4mol=> VCO2=0,4.22,4=8,96(L)
c,
có V không khí=11,2 L
oxi chiếm 1/5 thể tích không khí=> VO2=1/5.V(không khí)=1/5.11,2=2,24L
vậy thể tích oxi cần cho pư cháy là 2,24L
a)\(n_{Fe}=\dfrac{22,4}{56}=0,4mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,4 \(\dfrac{4}{15}\) \(\dfrac{2}{15}\)
\(V_{O_2}=\dfrac{4}{15}\cdot22,4=5,973l\)
b)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{8}{45}\) \(\dfrac{4}{15}\)
\(m_{KClO_3}=\dfrac{8}{45}\cdot122,5=21,78g\)
đốt cháy hoàn toàn 12g magie trong bình chứa khí oxi
\(a) 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ b) n_{O_2} = \dfrac{1}{2}n_{Mg} = \dfrac{1}{2}.\dfrac{12}{24} = 0,25(mol)\\ \Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)\\ V_{không\ khí} = 5V_{O_2} = 5,6.5 = 28(lít)\\ c) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5(mol)\\ m_{KMnO_4} = 0,5.158 =79(gam)\)