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a/ 2Al + 3S -to-> Al2S3
b/ Fe + S -to-> FeS
c/ Pb + S -to-> PbS
d/ 2Na + S -to-> Na2S
a/ 2Al + 3S -to-> Al2S3
b/ Fe + S -to-> FeS
c/ Pb + S -to-> PbS
d/ 2Na + S -to-> Na2S
3Fe+2O2 -t--> Fe3O4
2Na + Cl2 ---> 2NaCl
2K + S --> K2S
4P+ 5O2-t-> 2P2O5
a)
4Na + O2 -to--> 2Na2O (1)
3Fe + 2O2 --to--> Fe3O4 (2)
S + O2 --to--> SO2 (3)
CH4 +2O2 --to--> CO2 + 2H2O (4)
b) Pư hóa hợp: (1), (2), (3)
1) \(C+O2-->CO2\)==>Pư hóa hợp
\(2Mg+O2-->2MgO\)==>Pư hóa hợp
\(4Al+3O2-->2Al2O3\)==>Pư hóa hợp
\(C2H6+\frac{7}{2}O2-->2CO2+3H2O\)
\(C2H2+\frac{5}{2}O2-->2CO2+H2O\)
2)
\(3Fe+2O2-->Fe3O4\)
\(S+O2-->SO2\)
\(CH4+2O2-->CO2+2H2O\)
\(2Cu+O2-->2CuO\)
\(4P+5O2-->2P2O5\)
\(C3H8O+\frac{9}{2}O2-->3CO2+4H2O\)
\(C4H10+\frac{13}{2}O2-->4CO2+5H2O\)
\(C7H16+11O2-->7CO2+8H2O\)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\\ 4K+O_2\underrightarrow{t^o}2K_2O\\ Zn+O_2\underrightarrow{t^o}ZnO\\ S+O_2\underrightarrow{t^o}SO_2\\ 2S+3O_2\underrightarrow{t^o}2SO_3\\ 4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2C+O_2\underrightarrow{t^o}2CO\\ C+O_2\underrightarrow{t^o}CO_2\\ 2Ba+O_2\underrightarrow{t^o}2BaO\\ 2Fe+O_2\underrightarrow{t^o}2FeO\\ 4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\\ 3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(1,2H_2+O_2\underrightarrow{t}2H_2O\)
\(2Mg+O_2\underrightarrow{t}2MgO\)
\(2Cu+O_2\underrightarrow{t}2CuO\)
\(S+O_2\underrightarrow{t}SO_2\)
\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(C+O_2\underrightarrow{t}CO_2\)
\(4P+5O_2\underrightarrow{t}2P_2O_5\)
\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)
c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O2 phản ứng hết, Bài toán tính theo O2
\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)
\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)
\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)
\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)
\(4,a,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)
\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)
\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)
\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)
\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)
\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)
\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=0,46875\left(mol\right)\)
\(n_{SO_2}=0,3\left(mol\right)\)
Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.
\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)
\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)
\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)
\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)
\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)
\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)
\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)
\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)
\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)
\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)
Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.
\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)
\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)
$2CO+O_2\rightarrow 2CO_2$
$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$
$4Na+O_2\rightarrow 2Na_2O$
$2Zn+O_2\rightarrow 2ZnO$
$4Al+3O_2\rightarrow 2Al_2O_3$
Bài 1
H2+ 1/2O2 --> H2O
Mg + 1/2O2 --> MgO
Cu+ 1/2O2-->CuO
S+O2 -->SO2
4Al+ 3O2-->2Al2O3
C+ O2--> CO2
2P+5/2O2--> P2O5
Bài 2
CH4+2O2->CO2+2H2O
2C2H2+5O2->4CO2+2H2O
C2H6O+3O2->2CO2+3H2O
Bàu 1
a) 4P+5O2--->2P2O5
S+O2--->SO2
3Fe+2O2--->Fe3O4
C2H4+3O2-->2CO2+2H2O
4Na+O2--->2Na2O
trừ phản ứng C2H4 thì tất cả đề là phản ứng hóa hợp
Bài 2
2H2+O2--->2H2O
2Mg+O2--->2MgO
2Cu+O2--->2CuO
S+O2--->SO2
4Al+3O2--->2Al2O3
C+O2---->CO2
4P+5O2--->2P2O5
Bài 1 :
a,
\(4P+5O_2\underrightarrow{^{to}}2P_2O_5\) (1)
\(S+O_2\underrightarrow{^{to}}SO_2\)(2)
\(3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\)(3)
\(C_2H_4+3O_2\underrightarrow{^{to}}2CO_2\uparrow+2H_2O\)(4)
\(4Na+O_2\underrightarrow{^{to}}2Na_2O\)(5)
\(2Ca+O_2\underrightarrow{^{to}}2CaO\)(6)
b, PHản ứng hóa hợp : (1) ; (2) ; (3) ;(5) ; (6)
Bài 2 :
\(2H_2+O_2\underrightarrow{^{to}}2H_2O\)
\(2Mg+O_2\underrightarrow{^{to}}2MgO\)
\(2Cu+O_2\underrightarrow{^{to}}2CuO\)
\(S+O_2\underrightarrow{^{to}}SO_2\)
\(4Al+3O_2\underrightarrow{^{to}}2Al_3O_3\)
\(C+O_2\underrightarrow{^{to}}CO_2\)
\(4P+5O_2\underrightarrow{^{to}}2P_2O_5\)