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4Al+3O2-to>2Al2O3
0,4----0,3---------0,2 mol
n Al2O3=\(\dfrac{20,4}{102}\)=0,2 mol
=>m Al=0,4.27=10,8g
=>VO2=0,3.22,4=6,72l
=>Vkk=6,72.5=33,6l
4Al + 3O2 ---> 2Al2O3
0,4 0,3 0,2
nAl2O3 = 20,4 / 102 = 0,2 ( mol )
=> mAl = 0,4 . 27 = 10,8 (g)
V O2 = 0,3.22,4 = 6,72(l)
Vkk = 6,72 . 5 = 33,6(l)
\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{20,4}{102}=0,2mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,4 0,3 0,2 ( mol )
\(m_{Al}=n_{Al}.M_{Al}=0,4.27=10,8g\)
\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=6,72.5=33,6l\)
mol Al2O3=mA PTHH:Al l2O3/MAl2O3 =20.4÷(27×2+16×3)=0.2(mol)
PTHH:4Al+3O2--t°-->2Al2O3
mol--0.4----0.3-----------0.2
-->m Al phản ứng=nAl×MAl=0.2×27=5.4(g)
b, Vo2=no2×22.4=0.3×22.4=6.72(l)
--->Vkk cần dùng=6.72×100%÷20%=33.6(l)
Vậy.....
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)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)
\(m_{Mg}+m_{O_2}\rightarrow m_{MgO}\Leftrightarrow4,8g+m_{O_2}\rightarrow8\Leftrightarrow m_{O_2}=3,2g\)
a) PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)
b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\) \(\Rightarrow n_{O_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
c) PTHH: \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KClO_3}=0,1\cdot122,5=12,25\left(g\right)\)
a. \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
PTHH : 3Fe + 2O2 ---to---> Fe3O4
0,2 \(\dfrac{0.4}{3}\)
b. \(V_{O_2}=\dfrac{0.4}{3}.22,4=\dfrac{8.96}{3}\left(l\right)\)
c. PTHH : 2KClO3 -> 2KCl + 3O2
\(\dfrac{0.8}{3}\) \(\dfrac{0.4}{3}\)
\(m_{KClO_3}=\dfrac{0.8}{3}.122,5=\dfrac{98}{3}\left(g\right)\)
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 \(PTHH:n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}.1,5=1\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=n.M=1,122,5=122,5\left(g\right)\)
PTHH:
2Mg+O2=>2MgO
BTKL:mMg+mO2=mMgO
mMg+ 3,2 = 8
=>mMg =8-3,2=4,8g
\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\)
\(Áp.dụng.ĐLBTKL,ta.có:\\ m_{Mg}+m_{O_2}=m_{MgO}\\ \Rightarrow m_{Mg}=m_{MgO}-m_{O_2}=8-3,2=4,8\left(g\right)\)