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\(m_{Mg}+m_{O_2}\rightarrow m_{MgO}\Leftrightarrow4,8g+m_{O_2}\rightarrow8\Leftrightarrow m_{O_2}=3,2g\)
a)
\(n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_P = \dfrac{4}{5}n_{O_2} = 0,4(mol)\\ \Rightarrow m_P = 0,4.31 = 12,4(gam)\)
b)
\(n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\)
c)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)
nMg = 7,2/24 = 0,3 (mol)
PTHH: 2Mg + O2 -> (t°) 2MgO
Mol: 0,3 ---> 0,15 ---> 0,3
VO2 = 0,15 . 22,4 = 3,36 (l)
Vkk = 3,36 . 5 = 16,8 (l)
mMgO = 0,3 . 40 = 12 (g)
3Fe+2O2-to>Fe3O4
0,225--0,15
n Fe=\(\dfrac{12,6}{56}\)=0,225 mol
VO2=0,15.22,4=3,36l
2KClO3-to>2KCl+3O2
0,1---------------------0,15
=>m KClO3=0,1.122,5=12,25g
\(a,3Fe+2O_2\rightarrow Fe_3O_4\)
\(b,\)
Ta có : \(n_{Fe}=\dfrac{m}{M}=\dfrac{126}{56}=2,25\left(mol\right)\)
\(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.2,25=1,5\left(mol\right)\)
\(\Rightarrow VO_2=33,6\left(l\right)\)
\(c,\)
\(PTHH:2KClO_3\rightarrow2KCl+3O_2\)
Theo \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)
nFe = 33,6 : 56 = 0,6 (mol)
pthh : 3Fe + 2O2 -t--> Fe3O4
0,6--> 0,4------->0,2 (mol)
=> vO2 = 0,4.22,4 = 8,96 (mol)
=> mFe3O4 = 0,2.232 = 46,4 (g)
pthh : 2KClO3 -t--> 2KClO3 + 3O2
0,267<-----------------------0,4(mol)
mKClO3= 0,267 .122,5 = 32,67 (g)
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)
c, \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,3.232=69,6\left(g\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,4\left(mol\right)\Rightarrow m_{KClO_3}=0,4.122,5=49\left(g\right)\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{50,4}{56}=0,9\left(mol\right)\)
\(a.PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2 1
0,9 0,6 0,3
\(b.V_{O_2}=n.24,79=0,6.24,79=14,874\left(l\right)\)
\(c.m_{Fe_3O_4}=n.M=0,3.\left(56.3+16.4\right)=69,6\left(g\right)\)
\(d.V_{O_2}=14,874\left(l\right)\\ \Rightarrow n_{O_2}=\dfrac{V}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2 2 3
0,6 0,6 0,9
\(m_{KClO_3}=n.M=0,6.\left(39+35,5+16.3\right)=55,5\left(g\right).\)
PTHH:
2Mg+O2=>2MgO
BTKL:mMg+mO2=mMgO
mMg+ 3,2 = 8
=>mMg =8-3,2=4,8g
\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\)
\(Áp.dụng.ĐLBTKL,ta.có:\\ m_{Mg}+m_{O_2}=m_{MgO}\\ \Rightarrow m_{Mg}=m_{MgO}-m_{O_2}=8-3,2=4,8\left(g\right)\)