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ta có S chiếm 40%
=>m S=\(\dfrac{4.40}{100}\)=1,6g=>S=\(\dfrac{1,6}{32}\)=0,05 mol
=>m C=4-1,6=2,4g=>C=\(\dfrac{2,4}{12}\)=0,2 mol
C+O2-to>CO2
0,05-0,05
S+O2-to>SO2
0,2-0,2 mol
=>VO2=0,25.22,4=5,6l
\(a,m_C=10.36\%=3,6\left(kg\right)=3600\left(g\right)\\ n_C=\dfrac{3600}{12}=300\left(mol\right)\\ m_S=10-3,6=6,4\left(kg\right)=6400\left(g\right)\\ n_S=\dfrac{6400}{32}=200\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ n_{O_2\left(tổng\right)}=n_C+n_S=300+200=500\left(mol\right)\\ V_{O_2\left(tổng\right)\left(đktc\right)}=500.22,4=11200\left(l\right)\\ V_{kk}=\dfrac{100}{20}V_{O_2\left(tổng\right)\left(đktc\right)}=5.11200=56000\left(l\right)\\ b,V_{hh\left(CO_2,SO_2\left(đktc\right)\right)}=22,4.\left(n_C+n_S\right)=22,4.\left(300+200\right)=11200\left(l\right)\)
a)
4Al + 3O2 --to--> 2Al2O3
2Mg + O2 --to--> 2MgO
b) Gọi số mol Al, Mg là a, b (mol)
=> 27a + 24b = 7,8 (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
a--->0,75a----->0,5a
2Mg + O2 --to--> 2MgO
b--->0,5b------->b
=> 102.0,5a + 40b = 14,2
=> 51a + 40b = 14,2 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
nO2 = 0,75a + 0,5b = 0,2 (mol)
=> VO2 = 0,2.22,4 = 4,48 (l)
=> Vkk = 4,48 : 20% = 22,4 (l)
c)
mAl = 0,2.27 = 5,4 (g)
mMg = 0,1.24 = 2,4 (g)
a)
4Al + 3O2 --to--> 2Al2O3
2Mg + O2 --to--> 2MgO
b) Gọi số mol Al, Mg là a, b (mol)
=> 27a + 24b = 7,8 (1)
PTHH: 4Al + 3O2 --to--> 2Al2O3
a--->0,75a----->0,5a
2Mg + O2 --to--> 2MgO
b--->0,5b------->b
=> 102.0,5a + 40b = 14,2
=> 51a + 40b = 14,2 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
nO2 = 0,75a + 0,5b = 0,2 (mol)
=> VO2 = 0,2.22,4 = 4,48 (l)
=> Vkk = 4,48 : 20% = 22,4 (l)
c)
mAl = 0,2.27 = 5,4 (g)
mMg = 0,1.24 = 2,4 (g)
\(m_{Al}=27,8.19,2\%=5,4\left(g\right)\\ m_{Fe}=27,8-5,4=22,4\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\end{matrix}\right.\)
PTHH:
4Al + 3O2 --to--> 2Al2O3
0,2-->0,15------->0,1
3Fe + 2O2 --to--> Fe3O4
0,4-->4/15--------->2/15
\(\rightarrow\left\{{}\begin{matrix}V_{kk}=\left(0,15+\dfrac{4}{15}\right).22,4.5=\dfrac{140}{3}\left(l\right)\\m_{Cran}=0,1.102+\dfrac{2}{15}.232=\dfrac{617}{15}\left(g\right)\end{matrix}\right.\)
mAl=27,8.19,42%=5,4g
⇒nAl=\(\dfrac{5,4}{27}\)=0,2mol
⇒nFe=\(\dfrac{27,8-5,4}{56}\)=0,4mol
4Al+3O2to→2Al2O34
3Fe+2O2to→Fe3O4
⇒nO2=\(\dfrac{3}{4}\)nAl+\(\dfrac{2}{3}\)nFe=\(\dfrac{5}{12}\)mol
⇒Vkk=\(\dfrac{5}{12}\).22,4.5=46,67l
b,
mrắn=27,8+mO2=27,8+\(\dfrac{5}{12}\)32=41,1g
\(a,n_{Al}=\dfrac{19,2\%.27,8}{27}=\dfrac{1112}{5625}\left(mol\right)\\ n_{Fe}=\dfrac{\left(100\%-19,2\%\right).27,8}{56}=\dfrac{14039}{35000}\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{O_2\left(tổng\right)}=\dfrac{1112}{5625}.0,75+\dfrac{2}{3}.\dfrac{14039}{35000}\approx0,4156762\left(mol\right)\\ V_{kk\left(đktc\right)}\approx0,4156762.5.22,4\approx46,5557344\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{\dfrac{1112}{5625}}{2}=\dfrac{556}{5625}\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{\dfrac{14039}{35000}}{3}=\dfrac{14039}{105000}\left(mol\right)\\ m_{rắn}=\dfrac{14039}{105000}.232+\dfrac{556}{5625}.102=41,1016381\left(g\right)\)
mS = 4 . 40% = 1,6 (g)
mC = 4 - 1,6 = 2,4 (g)
nS = 1,6/32 = 0,05 (mol)
nC = 2,4/12 = 0,2 (mol)
PTHH: S + O2 -> (t°) SO2
Mol: 0,05 ---> 0,05
C + O2 -> (t°) CO2
Mol: 0,2 ---> 0,2
nO2 (cần dùng) = 0,05 + 0,2 = 0,25 (mol)
VO2 (cần dùng) = 0,25 . 22,4 = 5,6 (l)
Cs j đó chx đúm:)