pthh: Fe₂O₃ + 3H₂SO₄ ===> Fe₂(SO₄)₃ + 3H₂O

nFe₂O₃ = 8/160 = 0,05 (mol)

nH₂SO₄ = \(\dfrac{300\times9,8\%}{98}\)= 0,3 (mol)

ta thấy: \(\dfrac{0,05}{1}\)<\(\dfrac{0,3}{3}\) (mol)

⇒ Fe₂O₃ phản ứng hết, H₂SO₄ còn dư

theo pthh: nH₂SO₄ phản ứng = 3nFe₂O₃ phản ứng = 0,15 (mol)

nFe₂O₃ = nFe₂(SO₄)₃ = 0,05 (mol)

⇒ mFe₂(SO₄)₃ = 0,05 \(\times\) 400 = 20g

nH₂SO₄ dư = nH₂SO₄ ban đầu - nH₂SO₄ phản ứng = 0,3 -0,15 = 0,15 (mol)

mH₂SO₄ dư = 0,15 \(\times\) 98 = 14,7 g

mdd sau phản ứng = mFe₂O₃ + mdd H₂SO₄ = 300 + 8 = 308 (g)

C% Fe₂(SO₄)₃ = \(\dfrac{20}{308}\times100\%\) = 6,49 (%)

C% H₂SO₄ dư = \(\dfrac{14,7}{308}\times100\%\) = 4,77 (%)

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

                0,2------------------>0,4---->0,2

m_{dd sau pư} = 200 + 21,2 - 0,2.44 = 212,4(g)

=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)

$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$

$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$

Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$

$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$

$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$

\(n_{H_2}=0,9\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Mg}=b\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}65a+24b=40,65\\a+b=0,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4646341463\\b=0,4353658537\end{matrix}\right.\)

Xem lại đề nha em

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)

\(a,n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1      0,2             0,1          0,1

\(V_{H_2}=0,1.22,4=2,24l\\ b)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65g\)     

Bài 6:

\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)

PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

_______0,2________0,6______0,2 (mol)

a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)

b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)

Bài 7:

\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

______0,1______0,1_______0,1 (mol)

a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)

b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)

a,\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{HCl}=0,5.0,4=0,2\left(mol\right)\)

PTHH: 2K + 2H₂O ---> 2KOH + H₂ (Fe và Cu ko tan trong nước)

              0,2                                0,1

Fe + 2HCl ---> FeCl₂ + H₂ (Cu ko phản ứng với HCl)

0,1     0,2

m_{Chất rắn còn lại} = m_Cu = 6,6 (g)

\(\rightarrow\left\{{}\begin{matrix}m_K=39.0,2=7,8\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\\m_{Cu}=6,6\left(g\right)\end{matrix}\right.\)

\(\rightarrow m_{\text{hhkimloại}}=7,8+5,6+6,6=20\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}\%m_K=\dfrac{7,8}{20}=39\%\\\%m_{Fe}=\dfrac{5,6}{20}=28\%\\\%m_{Cu}=100\%-39\%-28\%=33\%\end{matrix}\right.\)

b, PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

\(n_{O\left(\text{trong oxit}\right)}=n_{H_2O}=n_{H_2}=0,1\left(mol\right)\\ \rightarrow n_{Fe\left(\text{trong oxit}\right)}=\dfrac{5,8-0,1.16}{56}=0,075\left(mol\right)\)

\(\rightarrow x:y=n_{Fe}:n_O=0,075:0,1=3:4\)

CTHH của oxit sắt Fe₃O₄

Sửa đề thành 2,24 l khí C nhé :)

 PTHH

Mg + 2HCl ----> MgCl2 + H2      (1)

MgO + 2HCl -----> MgCl2 + H2O (2)

a) Theo pt(1) n Mg = n H2 = \(\frac{1,12}{22,4}\) = 0,05 (mol)

==> m Mg = 0,005 . 24=1,2 (g)

%m Mg =  \(\frac{1,2}{3,2}\). 100%= 37,5%

%m MgO= 100% - 37,5%= 62,5%

b)m dd sau pư = 3,2 + 246,9 - 0,05 . 2=250 (g)

Theo pt(1)(2)  n MgCl2(1) = n Mg = 0,05 mol

                         n MgCl2 (2) = n MgO=\(\frac{3,2-1,2}{40}\)=0,05(mol)

==> tổng n MgCl2 = 0,1 (mol)  ---->m MgCl2 = 9,5 (g)

C%(MgCl2)= \(\frac{9,5}{250}\) .100% = 3,8%