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PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{15,8}{158}=0,1\left(mol\right)\)
a. Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}0,1=0,05\left(mol\right)\)
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,05.22,4=1,12\left(l\right)\)
b. PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Ta có: \(\dfrac{1}{n_{O_2}}=\dfrac{1}{0,05}\)
\(\dfrac{1}{n_{Fe}}=\dfrac{1}{0,1}\)
\(\Rightarrow\dfrac{1}{n_{O_2}}>\dfrac{1}{n_{Fe}}\)
Vậy Fe dư
Theo PTHH: \(n_{Fe_3O_4}=\dfrac{0,1.1}{3}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,73g\)
\(2KMnO_4\underrightarrow{to}K_2MnO_4+MnO_2+O_2\\ 3Fe+2O_2\underrightarrow{to}Fe_3O_4\\ n_{Fe_3O_4}=\dfrac{69,6}{232}=0,3\left(mol\right)\\ \Rightarrow n_{O_2}=2.0,3=0,6\left(mol\right)\\ n_{KMnO_4}=2.n_{O_2}=2.0,6=1,2\left(mol\right)\\ m=m_{KMnO_4}=158.1,2=189,6\left(g\right)\\ V=V_{O_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right)\)
\(n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ a,2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ 3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2(mol)\\ \Rightarrow V_{O_2}=0,2.22,4=4,48(l)\\ n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{15}(mol)\\ \Rightarrow m_{KClO_3}=\dfrac{2}{15}.122,5\approx 16,33(g)\)
\(n_{KMnO_4}=\dfrac{18.96}{158}=0.12\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
\(0.12...........................................0.06\)
\(V_{O_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.08.....0.06.......0.04\)
\(m_{Al\left(dư\right)}=\left(0.2-0.08\right)\cdot27=3.24\left(g\right)\)
\(m_{Al_2O_3}=0.04\cdot102=4.08\left(g\right)\)
\(n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2<---1,5---------->1
=> \(m_{Al_2O_3}=1.102=102\left(g\right)\)
\(m_{Al}=2.27=54\left(g\right)\)
a) \(n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,06->0,04------->0,02
=> mFe3O4 = 0,02.232 = 4,64 (g)
b) VO2 = 0,04.22,4 = 0,896 (l)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
0,15 0,1 0,05
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\
V_{O_2}=0,1.11,4=2,24\left(l\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
mik c.ơn trước
a)
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
b) $n_{KMnO_4} = \dfrac{79}{158} = 0,5(mol)$
Theo PTHH : $n_{O_2} = \dfrac{1}{2}n_{KMnO_4} = 0,25(mol)$
$\Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)$
c) $n_P = \dfrac{3,1}{31} = 0,1(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
Ta thấy : $n_P : 4 < n_{O_2} :5$ nên $O_2$ dư
$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,05(mol)$
$m_{P_2O_5} = 0,05.142 = 7,1(gam)$
a) \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,3<---0,2-------->0,1
=> m = 0,3.56 = 16,8 (g)
b) mFe3O4 = 0,1.232 = 23,2 (g)
c) Vkk = 4,48 : 20% = 22,4 (l)
nO2 = 4,48/22,4 = 0,2 (mol)
PTHH: 3Fe + 2O2 -> (t°) Fe3O4
Mol: 0,3 <--- 0,2 ---> 0,1
mFe = 0,3 . 56 = 16,8 (g)
mFe3O4 = 0,1 . 232 = 23,2 (g)
Vkk = 4,48 . 5 = 22,4 (l)
Câu 14 :
\(n_{KMnO4}=\frac{158}{158}=1\left(mol\right)\)
a. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
_____1_________0,5_________0,5_______0,5
Rắn gồm K2MnO4 và MnO2Ta có
\(m=197.0,5+87.0,5=142\left(g\right)\)
\(V_{O2}=0,5.22,4=11,2\left(l\right)\)
b. \(n_{Fe}=\frac{16,8}{56}=0,3\)
\(PTHH:3Fe+2O_2\rightarrow Fe_3O_4\)
Trước __0,3 ___0,5__________
Phản ứng__0,3___0,2____________
Sau______0___0,3______ 0,1
\(m_{Fe3O4}=0,1.232=23,2\left(g\right)\)
Câu 15 :
a) \(PTHH:2KClO_3\rightarrow2KCl+3O_2\)
\(n_{KClO3}=\frac{245}{122,5}=2\left(mol\right)\)
Cứ 2 mol KClO3 \(\rightarrow2\left(mol\right)KCl\rightarrow3\left(mol\right)O_2\)
____2 mol __________2 mol______ 3 mol
\(\rightarrow m=2.74,5=149\left(g\right)\)
\(\rightarrow V=3.22,4=67,2\left(l\right)\)
b) \(PTHH:2Mg+O_2\rightarrow2MgO\)
\(n_{Mg}=\frac{24}{24}=1\left(mol\right)\)
\(n_{O2}=\frac{67,2}{22,4}=3\left(mol\right)\)
Ta có: nMg/2 < nO2/1 nên O2 dư \(\rightarrow\) Tính theo Mg
\(\rightarrow n_{MgO}=\frac{1.2}{2}=1\left(mol\right)\)
\(\rightarrow m=2.40=40\left(g\right)\)