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\(1.2Fe+3Cl_2\overset{t^o}{--->}2FeCl_3\)
\(2.Zn+S\overset{t^o}{--->}ZnS\)
\(3.4P+5O_2\overset{t^o}{--->}2P_2O_5\)
\(4.Mg+2HCl--->MgCl_2+H_2\)
\(5.CO_2+H_2O--->H_2CO_3\)
\(6.K_2O+H_2O--->2KOH\)
\(7.4Na+O_2--->2Na_2O\)
\(8.Fe_2\left(SO_4\right)_3+3Ca\left(OH\right)_2--->2Fe\left(OH\right)_3\downarrow+3CaSO_4\)
\(9.Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
1) 2Fe+3Cl2 --to- > 2FeCl3
2)Zn+S --to- > ZnS
3) 4P+5O2 --to- > 2P2O5
4) Mg+ 3HCl ---> MgCl2 + H2
5)CO2+H2O --->H2CO3
6)K2O+H2O ----> 2KOH
7)4Na + O2 --to- > 2Na2O
8)Fe2(SO4)3 + 3Ca(OH)2 ----> 2Fe(OH)3+ 3CaSO4
9. Al2O3 + 3H2SO4 -----> Al2(SO4)3 + 3H2O
Lập các PTHH theo các sơ đồ phản ứng sau;
1. 2C + O2 → 2CO
2. 4Na + O2 → 2Na2O
3. Mg + 2HCl → MgCl2 + H2
4. Fe2O3 + 3H2 → 2Fe + 3H2O
5. Na2CO3 + 2HCl → 2NaCl + H2O + CO2
a/ 4Na + O2 ------> 2Na2O
b/ FeCl2 + 2KOH -------> Fe(OH)2 + 2KCl
c/ Mg + 2HCl -------> MgCl2 + H2
d/ 4P + 5O2 ------> 2P2O5
e/ AlCl3 + 3NaOH -------> Al(OH)3 + 3NaCl
f/ 2Al + 6HCl -------> 2AlCl3 + 3H2
Tui ghi hệ số thôi nha !
a/ 4 1 2
b/ 1 2 1 2
c/ 1 2 1 1
d/ 4 5 2
e/ 1 3 1 3
f/ 2 6 2 3
a) $4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
b) $Mg + Cl_2 \xrightarrow{t^o} MgCl_2$
c) $2Na + 2H_2O \to 2NaOH + H_2$
d) $C + O_2 \xrightarrow{t^o} CO_2$
e) $C_xH_y + (x + \dfrac{y}{4})O_2 \xrightarrow{t^o} xCO_2 + \dfrac{y}{2}H_2O$
f) $2Al + Fe_2O_3 \xrightarrow{t^o} Al_2O_3 + 2Fe$
g) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
i) $Fe_xO_y + yCO \xrightarrow{t^o} xFe + yCO_2$
k) $Fe_2O_3 + 6HCl \to 2FeCl_3 +3 H_2O$
l) $3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$
(*) 4K + O2 \(\rightarrow\) 2K2O
(*) Fe + Cl2 \(\rightarrow\) FeCl2 (phương trình cân bằng rồi nha)
(*) 2NaOH + MgCl2 \(\rightarrow\) Mg(OH)2 + 2NaCl
(*) Mg + 2HCl \(\rightarrow\) MgCl2 +H2
(*) 2Fe(OH)3 \(\rightarrow\) Fe2O3 + 3H2O
a) PTHH: 4K+O2→2K2O
Tỉ lệ: 4:1:2
b)2Al+3CuCl2→2AlCl3+3Cu
Tỉ lệ: 2:3:2:3
c)6NaOH+Fe2(SO4)3→2Fe(OH)3+3Na2SO4
Tỉ lệ: 6:1:2:3
#Walker
\(PTHH:4K+O2\rightarrow2K2O\)
\(4K:O2:2K2O=4:1:2\)
\(PTHH:2Al+3CuCl2\rightarrow2AlCl3+3Cu\)
\(2Al:3CuCl2:2AlCl2+3Cu=2:3:2:3\)
\(6NaOH+Fe2\left(SO4\right)3\rightarrow2Fe\left(OH\right)3+3Na2SO4\)
\(6NaOH:Fe2\left(SO4\right)3:2Fe\left(OH\right)3:3Na2SO4=6:1:2:3\)
a/ 4Na + O2 ===> 2Na2O
b/ 2Fe(OH)3 ==(nhiệt)==> Fe2O3 + 3H2O
c/ 2Al + 3H2SO4 ===> Al2(SO4)3 + 3H2
3.2/ mNaCl = 0,2 x 58,5 = 11,7 gam
3.3/ VCO2(đktc) = 1,25 x 22,4 = 28 lít
a) \(4Na+O_2\rightarrow2Na_2O\)
b)\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)
c)\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
3.2 \(m_{NaCl}=n.M=0,2.58,5=11,7\left(g\right)\)
3.3\(V_{CO_2}=n.22,4=1,25.22,4=28\left(lit\right)\)
a) Al2O3 + 6HCl → 2AlCl3 + 3H2O
b) CuO + H2 -to-➢ Cu + H2O
c) Mg + 2HCl → MgCl2 + H2
d) 4Na + O2 -to-➢ 2Na2O
a) Al2O3 + 6HCl → 2AlCl3↓ + 3H2O
b) CuO + H2 → Cu + H2O↑
c) Mg + 2HCl → MgCl2 + H2
d) 4Na + O2 → 2Na2O
\(a,4Na+O_2\xrightarrow{t^o}2Na_2O\\ b,Mg+2HCl\to MgCl_2+H_2\\ c,6NaOH+Fe_2(SO_4)_3\to 3Na_2SO_4+2Fe(OH)_3\downarrow\)
\(a.4Na+O_2-^{t^o}\rightarrow2Na_2O\\ b.Mg+2HCl\rightarrow MgCl_2+H_2\\ c.6NaOH+Fe_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Fe\left(OH\right)_3\)