PTHH : S + O₂ → SO₂

a) Áp dụng định luật bảo toàn khối lượng ta có:

\(m_S+m_{O_2}=m_{SO_2}\)

\(\Rightarrow m_{O_2}=m_{SO_2}-m_S=9,6-4,8=4,8\left(g\right)\)

b) \(n_{O_2}=\frac{m}{M}=\frac{4,8}{32}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=n.22,4=0,15.24=3,6\left(l\right)\)

c) \(V_{O_2}=V_{KK}.\frac{1}{5}\Rightarrow V_{KK}=V_{O_2}.5=3,6.5=18\left(l\right)\)

Do quặng chứa 10% tạp chất

=> FeS₂ chiếm 90%

\(m_{FeS_2}=\dfrac{125.90}{100}=112,5\left(g\right)\)

=> \(n_{FeS_2}=\dfrac{112,5}{120}=0,9375\left(mol\right)\)

PTHH: 4FeS₂ + 11O₂ --t^o--> 2Fe₂O₃ + 8SO₂

            0,9375---------------------->1,875

=> V_SO2 = 1,875.22,4 = 42 (l)

nAl2O3=0.2 mol 

4Al+3O2--->2Al2O3

0,2  0,2               0,2

mAl nguyên chất=0,2*27=5,4(g)

mAl=mAl nguyên chất+m tạp chất=5,4+ 5,4*15%=6,21(g)

b)Vo2=0,2*22,4=4,48=>Vk2 tự tính cho quen nhak

nFe = 46,4/56 = 29/35 (mol)

PTHH: 4Fe + 3O2 -> (t°) 2Fe2O3

Mol: 29/35 ---> 87/140 ---> 29/70

mFe2O3 = 29/70 . 160 = 464/7 (g)

Vkk = 87/140 . 5 . 22,4 = 69,6 (l)

Khối lượng C chứa trong 1 tấn than:

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

           12---32-------gam

           0,96--x-------tấn

\(\Rightarrow x=\dfrac{0,96.32}{12}=2,56\left(\text{tấn}\right)\)

\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ a,n_{O_2}=3.0,5=1,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\\ V_{kk\left(đktc\right)}=33,6.5=168\left(l\right)\\ b,n_{CO_2}=n_{H_2O}=2.0,5=1\left(mol\right)\\ m_{CO_2}=44.1=44\left(g\right);m_{H_2O}=18.1=18\left(g\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{Ca\left(OH\right)_2}=n_{CO_2}=1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=1.74=74\left(g\right)\\ m_{ddCa\left(OH\right)_2}=\dfrac{74.100}{10}=740\left(g\right)\)

\(a,n_{FeS_2}=\dfrac{m_{FeS_2}}{M_{FeS_2}}=\dfrac{6}{120}=0,05\left(mol\right)\\ 4FeS_2+11O_2\rightarrow\left(t^o,xt\right)2Fe_2O_3+8SO_2\uparrow\\ n_{Fe_2O_3}=\dfrac{2}{4}.n_{FeS_2}=\dfrac{2}{4}.0,05=0,025\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=160.0,025=4\left(g\right)\\ n_{SO_2}=\dfrac{8}{4}.n_{FeS_2}=\dfrac{8}{4}.0,05=0,1\left(mol\right)\\ \Rightarrow m_{SO_2}=0,1.64=6,4\left(g\right)\\ \Rightarrow m_{sp}=m_{Fe_2O_3}+m_{SO_2}=4+6,4=10,4\left(g\right)\\ b,n_{O_2}=\dfrac{11}{4}.n_{FeS_2}=\dfrac{11}{4}.0,05=0,1375\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,1375.22,4=3,08\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=3,08.5=15,4\left(l\right)\)

\(pthh:4FeS_2+11O_2\overset{t^o}{--->}2Fe_2O_3+8SO_2\uparrow\)

a. Ta có: \(n_{FeS_2}=\dfrac{6}{120}=0,05\left(mol\right)\)

Theo pt: \(n_{O_2}=\dfrac{11}{4}.n_{FeS_2}=\dfrac{11}{4}.0,05=0,1375\left(mol\right)\) 

\(\Rightarrow m_{sản.phẩm.thu.được}=6+0,1375.32=10,4\left(g\right)\)

b. Ta có: \(V_{O_2}=0,1375.22,4=3,08\left(lít\right)\)

Mà: \(V_{O_2}=\dfrac{1}{5}V_{kk}\)

\(\Rightarrow V_{kk}=3,08.5=15,4\left(lít\right)\)

a: \(4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\uparrow\)

\(n_{FeS_2}=\dfrac{6}{129}\)(mol)

\(\Leftrightarrow n_{Fe_2O_3}=\dfrac{3}{129}\left(mol\right)\)

\(m_{Fe_2O_3}=\dfrac{3}{129}\cdot178=\dfrac{178}{43}\left(g\right)\)

b: \(n_{O_2}=\dfrac{2}{129}\left(mol\right)\)

\(V=n\cdot22.4=\dfrac{2}{129}\cdot22.4=\dfrac{224}{645}\left(lít\right)\)

\(m_{FeS_2}=400.\left(100-10\right)\%=360\left(g\right)\\ \rightarrow n_{FeS_2}=\dfrac{360}{120}=3\left(mol\right)\)

PTHH: 4FeS₂ + 11O₂ --to--> 2Fe₂O₃ + 8SO₂

                3                                                3

\(\rightarrow V_{SO_2}=6.22,4=134,4\left(l\right)\)

Fe₂O₃ oxit bazo, SO₂ oxit axit

4P+5O2-to>2P2O5

0,4---0,5-----0,2

n P=0,4 mol

=>VO2=0,5.22,4=11,2l

=>Vkk=1,12.5=5,6l