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nMg = 9,6/24 = 0,4 (mol)
2Mg + O2 ---to---> 2MgO
0,4____0,2_________0,4
VO2(đktc) = 0,2.22,4 = 4,48(l)
mMgO = 0,4.40 = 16(g)
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, PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
b, Ta có: \(n_{MgO}=\dfrac{2,4}{40}=0,06\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,03\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,03.22,4=0,672\left(l\right)\)
c, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,02\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,02.122,5=2,45\left(g\right)\)
a: \(4Al+3O_2\rightarrow2Al_2O_3\)
c: \(n_{Al}=\dfrac{2.4}{24}=0.1\left(mol\right)\)
\(\Leftrightarrow n_{Al_2O_3}=0.05\left(mol\right)\)
\(m_{Al_2O_3}=0.05\cdot96=1.92\left(g\right)\)
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)\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
\(\dfrac{8}{45}\) \(\dfrac{4}{15}\)
\(m_{KClO_3}=\dfrac{8}{45}\cdot122,5=21,78g\)
a)\(2Mg + O_2 \xrightarrow{t^o} 2MgO\)
b)
\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)
Theo PTHH :
\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)
c)
\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)
d)
\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)
a)
PTHH: 3Fe + 2O2 ____\(t^o\)____> Fe3O4 (1)
b) Ta có: nFe = \(\dfrac{25.2}{56}=0.45\left(mol\right)\)
Theo (1): n\(O_2\)= \(\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}0.45=0.3\left(mol\right)\)
=> \(V_{O_2\left(đktc\right)}=0.3\cdot22.4=6.72\left(l\right)\)
c) PTHH: 2KClO3 __\(t^o\)___> 2KCl + 3O2 (2)
-Muốn điều chế được lượng oxi dùng cho phản ứng trên thì \(n_{O_2\left(2\right)}=n_{O_2\left(1\right)}=0.3\left(mol\right)\)
Theo (2) \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}0.3=0.2\left(mol\right)\)
=> \(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)
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)2Mg+O2-----t0-->(ko có dấu gạch đứt nha bạn) 2MgO
1 1 (mol)
0,1 -> 0,05(mol)
b)nMg(Mg viết xuống dưới n dùm mình)=2,4/24=0,1 mol
=>Vo2=n.22,4=0,05.22,4=1,12 lít
Mình cảmm mơn ạ