Cho 24g Sắt (III) oxit và 300g dung dịch H2SO4 19,6% sau khi phản ứng xong thì:

a) thu được mấy gam muối?

b) dung dịch sau khi phản ứng chứa chất gì? Nồng độ mấy %?

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a) nFe2O3= 24/160= 0,15(mol)

mH2SO4= 300.19,6%=58,8(g) -> nFe2O3= 58,8/98= 0,6(mol)

PTHH: Fe2O3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O

Ta có: 0,15/1 < 0,6/3

-> Fe2O3 p.ứ hết, H2SO4 dư. Tính theo nFe2O3.

=> nFe2(SO4)3= nFe2O3= 0,15(mol)

=> m(muối)= mFe2(SO4)3= 400.0,15= 60(g)

b) Dung dịch sau p.ứ gồm Fe2(SO4)3 và H2SO4(dư).

mH2SO4(dư)= (0,6-0,15.3).98=14,7(g)

mddsau=mddH2SO4 + mFe2O3 = 300+24= 324(g)

=> \(C\%ddFe2\left(SO4\right)3=\frac{60}{324}.100\approx18,519\%\\ C\%ddH2SO4\left(dư\right)=\frac{14,7}{324}.100\approx4,537\%\)

\(a.n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.80}{36,5}=0,32\left(mol\right)\\ a.Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,05}{1}< \dfrac{0,32}{2}\\ \Rightarrow HCldư\\ \Rightarrow n_{H_2}=n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b.Chất.trong.dd.sau.phản.ứng:FeCl_2,HCl\left(dư\right)\\ m_{ddsau}=2,8+80-0,05.2=82,7\left(g\right)\\ C\%_{ddFeCl_2}=\dfrac{0,05.127}{82,7}.100\approx7,678\%\\n_{HCl\left(dư\right)}=0,32-0,05.2=0,22\left(mol\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,22.36,5}{82,7}.100\approx9,71\% \)

a,\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂SO₄ →CuSO₄ + H₂O

Mol:     0,25      0,25         0,25

\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{19,6}=125\left(g\right)\)

b,m_{dd sau pứ} = 20+125 = 145 (g)

\(C\%_{ddCuSO_4}=\dfrac{0,25.160.100\%}{145}=27,59\%\)

\(Cu+H_2SO_4\rightarrow CuSO_4+H_2\)
0,3125    0,3125   0,3125        (mol)
a)\(n_{Cu}=\dfrac{20}{64}=0,3125\left(mol\right)\)
\(m_{H_2SO_4}=0,3125.98=30,625\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{30,625}{19,6}.100=156,25\left(g\right)\)
b)\(m_{CuSO_4}=0,3125.160=50\left(g\right)\)
\(m_{ddCuSO_4}=20+156,25=176,25\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{50}{176,25}.100\approx28,37\%\)

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ a.0,25.........0,25..........0,25.......0,25\left(mol\right)\\ Đặt:a=m_{ddH_2SO_4}\left(g\right)\Rightarrow m_{ddX}=a+20\left(g\right)\\Vì.axit.dư\Rightarrow m_{CuSO_4}=0,25.160=40\left(g\right)\\ \dfrac{40}{a+20}.100\%=16\%\Leftrightarrow a=230\left(g\right)\\ \Rightarrow m_{ddH_2SO_4}=230\left(g\right)\\ b.m_{H_2SO_4\left(dư\right)}=230.19,6\%-0,25.98=20,58\left(g\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{20,58}{230+20}.100=8,232\%\)

nZn = 19,5/65 = 0,3 (mol)

nH2SO4 = 49/98 = 0,5 (mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

LTL: 0,3 < 0,5 => H2SO4 dư

nH2SO4 (phản ứng) = mZnSO4 = 0,3 (mol)

mH2SO4 (dư) = (0,5 - 0,3) . 98 = 19,6 (g)

mZnSO4 = 0,3 . 161 = 48,3 (g)

PTHH: CuO + H2 -> (t°) Cu + H2O

Mol: 0,3 <--- 0,3 ---> 0,3

mCu = 0,3 . 64 = 19,2 (g)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=n_{Fe(OH)_2}=0,2(mol)\\ a,m_{Fe}=0,2.56=11,2(g)\\ b,C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1M\)

\(c,Ba(OH)_2+FeSO_4\to BaSO_4\downarrow+Fe(OH)_2\downarrow\\ n_{Ba(OH)_2}=\dfrac{250.17,1}{100.171}=0,25(mol)\\ LTL:\dfrac{0,2}{1}<\dfrac{0,25}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=233.0,2=46,6(g)\)

Đáp án : A

Đáp án A

nFe = 2. 58/400 = 0,29 mol

nFeSO₄ = x , nFe₂(SO₄)₃ = y

⇒mdd = 152 x + 400y = 51,76

nFe = nFeSO₄ +2.nFe₂(SO₄)₃  = x+ 2y =  0,29

⇒x= 0,13 mol , y= 0,08 mol

BT S: nH₂SO₄ = nFeSO₄ + 3Fe₂(SO₄)₃ = 0,13 + 3.0,08 = 0,37

⇒b = 0,37.98/9,8% = 370g