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\(n_{FeO}=\dfrac{7,2}{72}=0,1mol\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
0,1 0,1 0,1
\(m_{H_2SO_4}=0,1\cdot98=9,8g\)
\(m_{ddH_2SO_4}=\dfrac{9,8}{24,5\%}\cdot100\%=40g\)
\(m_{FeSO_4}=0,1\cdot152=15,2g\)
\(m_{ddsau}=7,2+40=47,2g\)
\(n_{FeSO_4.7H_2O}=a\left(mol\right)\Rightarrow m=278a\left(g\right)\)
\(m_{FeSO_4còn}=15,2-152a\left(g\right)\)
Dung dịch sau khi làm lạnh có khối lượng:
\(m_{ddsaull}=47,2-278a\left(g\right)\)
\(\Rightarrow C\%=\dfrac{15,2-152a}{47,2-278a}\cdot100\%=13,6\%\Rightarrow a=0,08mol\)
\(\Rightarrow m=278a=278\cdot0,08=22,24g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mCuSO4 trong tinh thể = \(\dfrac{160m}{250}\)= 0,64m
=> mH2O trong tinh thể = 0,36m
mH2O còn sau khi tách tinh thể = 152,25 - 0,36m
m CuSO4 trong dd bảo hoà = 35,5 - 0,64m = 0,207.(152,25 - 0,36m)
=> m =7,05(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2RS+3O_2\underrightarrow{^{^{t^0}}}2RO+2SO_2\)
\(RO+H_2SO_4\rightarrow RSO_4+H_2O\)
Giả sử :
\(n_{H_2SO_4}=1\left(mol\right)\)
\(\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{98}{24.5\%}=400\left(g\right)\)
\(m_{\text{dung dịch muối}}=R+16+400=R+416\left(g\right)\)
\(C\%_{RSO_4}=\dfrac{R+96}{R+416}\cdot100\%=33.33\%\)
\(\Rightarrow R=64\)
\(R:Cu\)
\(n_{CuS}=\dfrac{12}{96}=0.125\left(mol\right)\)
\(n_{CuSO_4}=n_{CuS}=0.125\left(mol\right)\)
\(m_{CuSO_4}=0.125\cdot160=20\left(g\right)\)
\(m_{dd}=0.125\cdot80+\dfrac{0.125\cdot98}{24.5\%}=60\left(g\right)\)
Khối lượng dung dịch bão hòa còn lại :
\(60-15.625=44.375\left(g\right)\)
\(CT:CuSO_4\cdot nH_2O\)
\(m_{CuSO_4}=m\left(g\right)\)
\(C\%=\dfrac{m}{44.375}\cdot100\%=22.54\%\)
\(\Rightarrow m=10\)
\(m_{CuSO_4\left(tt\right)}=20-10=10\left(g\right)\)
\(\dfrac{10}{15.625}=\dfrac{160}{M_{tt}}\)
\(\Rightarrow M_{tt}=250\)
\(\Rightarrow n=5\)
\(CT:CuSO_4\cdot5H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{BaSO_4}=\dfrac{23.3}{233}=0.1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+H_2O\)
\(n_{BaO}=n_{Ba\left(OH\right)_2}=n_{BaSO_4}=0.1\left(mol\right)\)
\(m_{BaO}=0.1\cdot153=15.3\left(g\right)\)
\(m_{Na_2O}=24.6-15.3=9.3\left(g\right)\)
\(n_{Na_2O}=\dfrac{9.3}{62}=0.15\left(mol\right)\)
\(\%BaO=62.2\%\)
\(\%Na_2O=37.8\%\)
\(2.\)
\(m_{ddX}=24.6+73.7=98.3\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{0.15}{2}+0.1=0.175\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{0.175\cdot98\cdot100}{19.6}=87.5\left(g\right)\)
\(m_{ddY}=m_{ddX}+m_{ddH_2SO_4}-m_{\downarrow}=98.3+87.5-23.3=162.5\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{0.075\cdot142}{162.5}\cdot100\%=6.55\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mH2SO4= \(\dfrac{300.7,35}{100}=22,05g\)
nH2SO4= \(\dfrac{22,05}{98}=0,225 mol\)
mHCl= \(\dfrac{200.7,3}{100}=14,6g\)
nHCl= \(\dfrac{14,6}{36,5}=0,4mol\)
H2SO4 + 2HCl → 2H2O + Cl2 ↑+ SO2 ↑
n trước pư 0,225 0,4
n pư 0,2 ← 0,4 → 0,4 → 0,2 → 0,2 mol
n sau pư dư 0,025 hết
a) mCl2= 0,2. 71= 14,2g
mSO2= 64. 0,2= 12,8g
mH2O= 18. 0,4=7,2g
mdd sau pư= 300 +200 -14,2 -12,8= 473g
C%dd H2O= \(\dfrac{7,2.100}{473}=1,52\)%
b) Mg + 2H2O → Mg(OH)2 + H2 ↑
x → 2x → x → x
Fe + 2H2O → Fe(OH)2 + H2↑
y → 2y → y → y
Gọi x,y lần lượt là số mol của Mg,Fe.
Ta có hệ phương trình:
24x + 56y = 8,7 x= \(\dfrac{5}{64}\)
⇒
2x + 2y = 0,4 y= \(\dfrac{39}{320}\)
VH2= 22,4. \((\dfrac{5}{64}+\dfrac{39}{320})\)= 4,48l
mhh MG(OH)2, Fe(OH)2= 8,7 +250 - 2.(\(\dfrac{5}{64}+\dfrac{39}{320}\)) = 2258,3g
mMg=24. \(\dfrac{5}{64}\)=1.875g
mFe= 8,7-1,875= 6,825g
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1........0.3.........0.1...........0.15\)
\(m_{dd_{HCl}}=97.8\cdot1=97.8\left(g\right)\)
\(m_{ddsaupư}=2.7+97.8-0.15\cdot2=100.2\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{100.2}\cdot100\%=13.32\%\)