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9. M + 2nHNO3 → M(NO3)n + nNO2 + nH2O
x1 | \(M\rightarrow M^{n+}+ne\)
xn | \(N^{+5}+1e\rightarrow N^{+4}\)
10. 3M + 4nHNO3 → 3M(NO3)n + nNO + 2nH2O
x3 | \(M\rightarrow M^{n+}+ne\)
xn | \(N^{+5}+3e\rightarrow N^{+2}\)
11. 8M + 10nHNO3 → 8M(NO3)n + nN2O + 5nH2O
x8 | \(M\rightarrow M^{n+}+ne\)
xn | \(2N^{+5}+8e\rightarrow N_2^{+1}\)
12. 10M + 12nHNO3 → 10M(NO3)n +n N2 + 6nH2O
x10 | \(M\rightarrow M^{n+}+ne\)
xn | \(2N^{+5}+10e\rightarrow N_2^0\)
13. 8M + 10nHNO3 → 8M(NO3)n + nNH4NO3 + 3nH2O
x8 | \(M\rightarrow M^{n+}+ne\)
xn | \(N^{+5}+8e\rightarrow N^{-3}\)
14. 3Fe3O4 + 28HNO3loãng → 9Fe(NO3)3 + NO + 14H2O
15. 3FexOy + (12x - 2y)HNO3 \(\rightarrow\) 3xFe(NO3)3 + (3x - 2y)NO + (6x - y)H2O
x(3x-2y) | Fex+2y/x → xFe+3 + (3x-2y) e
x3 | N+5 + 3e → N+2
16.(5x - 2y)Mg + (12x - 4y)HNO3 \(\rightarrow\) (5x - 2y)Mg(NO3)2 + 2NxOy + (6x - 2y)H2O
x(5x-2y) | \(Mg\rightarrow Mg^{2+}+2e\)
x2 | \(xN^{+5}+\left(5x-2y\right)e\rightarrow N^{\dfrac{2y}{x}}_x\)
17. 10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O
x5 | 2Fe+2→Fe2+3 + 2e
x2 | Mn+7+ 5e → Mn+2
18. 5SO2 + 2KMnO4 + 2H2O → 2MnSO4 + K2SO4 + 2H2SO4
x5 | S+4 → S+6 + 2e
x2 | Mn +7 + 5e → Mn+2
19. SO2 + Br2 + 2H2O → H2SO4 + 2HBr
x1 | S+4 → S+6 + 2e
x1 | Br20 + 2e → 2Br-
20 . 4FeS2 + 11O2 → 2Fe2O3 + 8SO2
x2 | 2FeS20 → Fe2+3 + 4S+4 + 22e
x11 | O20 +4e → 2O2-
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2) nH2=0,1(mol)
a) PTHH: Fe +2 HCl -> FeCl2 + H2
0,1______0,2______0,1____0,1(mol)
nFe=nH2=0,1(mol)
=>mFe=nFe.M(Fe)=0,1.56=5,6(g)
=> mFeO=mX-mFe= 9,2-5,6=3,6(g)
=> nFeO=mFeO/M(FeO)=3,6/72=0,05(mol)
PTHH: FeO +2 HCl -> FeCl2 + H2
0,05_________0,1___0,05__0,05(mol)
b) Sao lại mỗi oxit a, có một oxit thôi mà :( Chắc % KL mỗi chất.
%mFeO=(mFeO/mhh).100%=(3,6/9,2).100=39,13%
=>%mFe=100%-%mFeO=100%-39,13%=60,87%
c) nHCl(tổng)= 2.nFe +2.nFeO=2.0,1+2.0,05=0,3(mol)
=>mHCl=nHCl.M(HCl)=0,3.36,5=10,95(g)
=>mddHCl=(mHCl.100%/C%ddHCl=(10,95.100)/7,3=150(g)
d) - Dung dich thu được chứa FeCl2.
mFeCl2=nFeCl2(tổng) . M(FeCl2)= (0,1+0,05).127=19,05(g)
mddFeCl2=mddHCl+mhh-mH2=150+9,2-0,1.2=159(g)
=> C%ddFeCl2=(mFeCl2/mddFeCl2).100%=(19,05/159).100=11,981%
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a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,2<-----------------0,2<--0,2
=> \(\%Mg=\dfrac{0,2.24}{10}.100\%=48\%\)
b) \(C\%\left(MgCl_2\right)=\dfrac{0,2.95}{0,2.24+200-0,2.2}.100\%=9,3\%\)
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Ta có: nNaOH=20/40=0,5(mol)
nCO2=22/44=0,5(mol)
Ta có: 0,5/0,5=1
=> Chỉ tạo một muối axit duy nhất NaHCO3.
PTHH: NaOH + CO2 -> NaHCO3
Ta có: nNaHCO3=nNaOH=0,5(mol)
=>mNaHCO3=0,5. 84=42(g)
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\)
Có \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,5}{0,5}=\dfrac{1}{1}\)
\(\Rightarrow\) Phản ứng tạo muối axit.
PTHH \(NaOH+CO_2\rightarrow NaHCO_3\)
\(n_{NaHCO_3}=n_{NaOH}=0,5\left(mol\right)\)
\(m_{NaHCO_3}=0,5\cdot84=42\left(g\right)\)
1. 2FeO + 4H2SO4 đặc ,nóng → Fe2(SO4)3 + SO2 +4H2O
x1 | \(2Fe^{+2}\rightarrow Fe^{+3}_2+2e\)
x1 | \(S^{+6}+2e\rightarrow S^{+4}\)
2. 2Fe3O4 + 10H2SO4 đặc ,nóng → 3Fe2(SO4)3 + SO2 + 10H2O
x1 | 2Fe3+8/3 \(\rightarrow\) 3Fe2+3 + 2e
x1 | \(S^{+6}+2e\rightarrow S^{+4}\)
3. 2Fe(OH)2 + 4H2SO4 → Fe2(SO4)3 + SO2 + 6H2O
x1 | \(2Fe^{+2}\rightarrow Fe^{+3}_2+2e\)
x1 | \(S^{+6}+2e\rightarrow S^{+4}\)
4. 2FeCO3 + 4H2SO4 → Fe2(SO4)3 + SO2 + 2CO2 + 4H2O
x1 | \(2Fe^{+2}\rightarrow Fe^{+3}_2+2e\)
x1 | \(S^{+6}+2e\rightarrow S^{+4}\)
5. 2FexOy + (6x-2y)H2SO4 \(\rightarrow\) xFe2(SO4)3 + (3x-2y)SO2 + (6x-2y)H2O
x1 | 2Fex+2y/x \(\rightarrow\) xFe2+3 +(6x-4y)e
x(3x-2y) | \(S^{+6}+2e\rightarrow S^{+4}\)