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nFeCl2 = CM.V = 0,15.0,2 = 0,03 mol
PTHH:
FeCl2 + 2NaOH → Fe(OH)2 + 2NaCl
0,03 → 0,06 → 0,03 → 0,06 (mol)
4Fe(OH)2 + O2 --to--> 2Fe2O3 + 4H2O
0,03 → 0,015
Chất rắn thu được sau khi nung kết tủa tới khối lượng không đổi là Fe2O3
→ m = mFe2O3 = 0,015.160 = 2,4 (g)
Dung dịch sau khi lọc kết tủa chỉ chứa 0,06 mol NaCl và có thể tích là V dd sau pư = 0,2 + 0,3 = 0,5 lít
→ CM NaCl = n/V = 0,06 / 0,5 = 0,12M
c2
a) 2KOH+H2SO4--->K2SO4+2H2O
m H2SO4=200.14,7/100=29,4(g)
n H2SO4=29,4/98=0,3(mol)
n KOH=2n H2SO4=0,6(mol)
m KOH=0,6.56=33,6(g)
m dd KOH=33,6.100/5,6=600(g)
V KOH=600/10,45=57,42(ml)
b) m dd sau pư=600+200=800(g)
n K2SO4=n H2SO4=0,3(mol)
m K2SO4=174.0,3=52,2(g)
C% K2SO4=52,2/800.100%=6,525%
c3
nCuO=3,2:80=0,04 mol
PTHH: CuO+H2SO4=>CuSO4+H2O
0,04mol->0,04mol->0,04mol->0,04mol
=> m H2SO4=0,04.98=3,92g
=> m ddH2SO4 tham gia phản ứng =3,92.100\4,9=80g
theo địnhluật bảo toàn khối lượng => m CuSO4= mCuO+mH2SO4-mH2O=3,2+80-0,04.18=82,48g
m CuSO4 thu được= 0,04.160=6,4g
=> C% CuSO4 =6,4\82,48.100=7,76%
a, \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
b, \(n_{FeCl_3}=0,4.2=0,8\left(mol\right)\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=2,4\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{2,4}{0,4+0,2}=4\left(M\right)\)
c, \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=\dfrac{1}{2}n_{FeCl_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,4.160=64\left(g\right)\)
\(n_{FeCl3}=2.0,4=0,8\left(mol\right)\)
PTHH : \(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
0,8----------------------->0,8----------->2,4
b) \(C_{MNaCl}=\dfrac{2,4}{0,4+0,2}=4M\)
c) \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)
0,8--------------->0,4
\(\Rightarrow a=m_{Fe2O3}=0,4.160=64\left(g\right)\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
nCO2 = 0,3 mol
CO2 + Ba(OH)2 -> BaCO3 + H2O
0,3..........0,3...............0,3
mBaCO3 = 0,3 .197 = 59,1 g
CM Ba(OH)2 = 0,3 / 0,6 = 0,5 M
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)
=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)
\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)
\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)
=> m Fe2O3 = 0,1 . 160=16(g)
\(n_{FeCl_3}=0.2\cdot0.4=0.08\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(0.08...........0.24..............0.08\)
\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)
\(0.08...........0.04\)
\(m_{Fe_2O_3}=0.04\cdot160=6.4\left(g\right)\)
\(V_{dd_{NaOH}}=\dfrac{0.24}{0.5}=0.48\left(l\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\) (1)
\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\) (2)
\(n_{FeCl_3}=0,2.0,4=0,08\left(mol\right)\)
Bảo toàn nguyên tố Fe : \(n_{FeCl_3}=2n_{Fe_2O_3}=0,08\left(mol\right)\)
=> \(n_{Fe_2O_3}=0,04\left(mol\right)\)
=> \(m_{Fe_2O_3}=0,04.160=6,4\left(g\right)\)
Theo PT (1) : \(n_{NaOH}=3n_{FeCl_3}=0,08.3=0,24\left(mol\right)\)
=> \(V_{NaOH}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)
a)FeCl3+3NaOH--->3NaCl+Fe(OH)3
2Fe(OH)3--->Fe2O3+3H2O(2)
b)n FeCl3=0,15.0,2=0,03(mol)
Theo pthh1
n Fe(OH)3=n FeCl3=0,03(mol)
Theo pthh2
n Fe2O3=1/2n Fe(OH)3 =0,015(mol)
m=m Fe2O3=0,015.160=2,4(g)
c)Theo pthh
n NaCl=3n FeCl3=0,09(mol)
CM NaCl=0,09/0,3=0,3(M)
Câu 2.
a) 2KOH+H2SO4--->K2SO4+2H2O
m H2SO4=200.14,7.100%=29,4(g)
n H2SO4=29,4/98=0,3(mol)
Theo pthh
n KOH=2n H2SO4=0,6(mol)
m KOH=0,6.56=33,6(g)
m dd KOH=33,6.100/5,6=600(g)
V KOH=600/10,45=57,41(ml)
m dd sau pư=600+200=800(g)
n K2SO4=n H2SO4=0,3(mol)
m K2SO4=0,3.174=52,2(g)
C% K2SO4=52,2/800.100%=6,525%
Bài 3
a) CuO+H2SO4--->CuSO4+H2O
n CuO=3,2/80=0,04(mol)
Theo pthh
n H2SO4=n CUSO4=n CuO=0,04(mol)
m H2SO4=0,04.98=3,92(g)
m dd H2SO4=3,92.100/4,9=80(g)
m CuSO4=0,04.160=6,4(g)
m dd sau pư=3,2+80=83,2(g)
C% CuSO4=6,4/83,2.100%=7,7%