Bài 3 Tính % a)Cu SO4 b) Ca (NO3) c)Al2(SO4)²3
K
Khách
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H
27 tháng 8 2021
$a) CaO + H_2O \to Ca(OH)_2$
$b) NaOH + HCl \to NaCl + H_2O$
$c) Ca(OH)_2 + 2HNO_3 \to Ca(NO_3)_2 + 2H_2O$
$d) AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$
$e) Mg(NO_3)_2 + Ca(OH)_2 \to Ca(NO_3)_2 + Mg(OH)_2$
$f) CuCl_2 + 2KOH \to Cu(OH)_2 + 2KCl$
$g) Fe(OH)_2 + H_2SO_4 \to FeSO_4 + 2H_2O$
$h) 2Fe(OH)_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 6H_2O$
$i) Al_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2AlCl_3$
21 tháng 12 2021
a) 3Fe +2 O2 --to--> Fe3O4
b) 4P + 5O2 --to--> 2P2O5
c) Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
d) Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
e) 2Cu(NO3)2 --to--> 2CuO + 4NO2 + O2
21 tháng 12 2021
\(a,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ c,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ d,Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ e,2Cu(NO_3)_2\xrightarrow{t^o}2CuO+4NO_2+O_2\uparrow\)
ND
Nguyễn Đức Trí
VIP
16 tháng 7 2023
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)
\(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)
16 tháng 7 2023
1. Fe + 2HCl → FeCl2 + H2
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
CuO + H2SO4 → CuSO4 + H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
CuO + 2HNO3 → Cu(NO3)2 + H2O
TT
2
HP
24 tháng 10 2021
\(PTK_{Mg\left(OH\right)_2}=24+\left(16+1\right).2=58\left(đvC\right)\)
\(PTK_{Ca\left(H_2PO_4\right)_2}=40+\left(1.2+31+16.4\right).2=234\left(đvC\right)\)
\(PTK_{Ba_3\left(PO_4\right)_2}=137.3+\left(31+16.4\right).2=601\left(đvC\right)\)
\(PTK_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)
\(PTK_{Ca\left(HCO_3\right)_2}=40+\left(1+12+16.3\right).2=162\left(đvC\right)\)
\(PTK_{Fe\left(NO_3\right)_2}=56+\left(14+16.3\right).2=180\left(đvC\right)\)
24 tháng 10 2021
\(PTK_{Mg\left(OH\right)_2}=24+\left(16+1\right)\cdot2=58\left(đvC\right)\\ PTK_{Ca\left(H_2PO_4\right)_2}=40+\left(2+31+16\cdot4\right)\cdot2=234\left(đvC\right)\\ PTK_{Ba_3\left(PO_4\right)_2}=137\cdot3+\left(31+16\cdot4\right)\cdot2=601\left(đvC\right)\\ PTK_{Al_2\left(SO_4\right)_3}=27\cdot2+\left(32+16\cdot4\right)\cdot3=342\left(đvC\right)\\ PTK_{Ca\left(HCO_3\right)_2}=40+\left(1+12+16\cdot3\right)\cdot2=162\left(đvC\right)\\ PTK_{Fe\left(NO_3\right)_2}=56+\left(14+16\cdot3\right)\cdot2=180\left(đvC\right)\)
TV
15 tháng 8 2016
3Cu+ 8HNO3--> 3Cu(NO3)2+ 2NO+ 4H2O
Cu+ 4HNO3-->Cu(NO3)2+2NO2+ 2H2O
2Fe+ 6H2SO4-->Fe2(SO4)3+ 3SO2+ 6H20
2Al+ 6H2SO4--> Al2(SO4)3+ 3SO2+ 6H2O
6Al+ 12H2SO4--> 3Al2(SO4)3+3S+ 12H2O
3Al+ 12HNO3--> 3Al(NO3)3+ 3NO+ 6H2O
10Al+ 32HNO3--> 10Al(NO3)3+ N2+ 16H2O
8Al+ 30HNO3--> 8Al(NO3)3+ 3NH4NO3+ 9H2O
TP
12 tháng 8 2021
1.2FeO + \(\dfrac{1}{2}\)O2 → Fe2O3
2.2FexOy + \(\dfrac{3x-2y}{2}\) O2 → xFe2O3
3.2Al + 6HCl → 2AlCl3 + 3H2
4.Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O
5.Al(OH)3 + 3HNO3 → Al(NO3)3 + 3H2O
TP
12 tháng 8 2021
6. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
7.2C4H10 + 13O2 → 8CO2 + 10H2O
8. 2KMnO4 → K2MnO4 + MnO2 + O2
9.2Cu(NO3)2 → 2CuO + 4NO2 + O2
10.M2Ox + 2xHNO3 → 2M(NO3)x + xH2O
NT
1
`a,`
\(K.L.P.T_{CuSO_{\text{4}}}=64+32+16.4=160< amu>.\)
\(\%Cu=\dfrac{64.100}{160}=40\%\)
\(\%S=\dfrac{32.100}{160}=20\%\)
\(\%O=100\%-40\%-20\%=40\%\)
`b,` \(K.L.P.T_{Ca\left(NO_3\right)}=40+14+16.3=102< amu>.\)
\(\%Ca=\dfrac{40.100}{102}\approx39,22\%\)
\(\%N=\dfrac{14.100}{102}\approx13,73\%\)
\(\%O=100\%-39,22\%-13,73\%=47,05\%\)
`c,` (bạn sửa lại đề: CTHH là `Al_2(SO_4)_3`)
\(K.L.P.T_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342< amu>.\)
\(\%Al=\dfrac{27.2.100}{342}\approx15,79\%\)
\(\%S=\dfrac{32.3.100}{342}\approx28,07\%\)
\(\%O=100\%-15,79\%-28,07\%=56,04\%\)