PTHH: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{40}{40}=1\left(mol\right)\\n_{H_2SO_4}=\dfrac{300\cdot98\%}{98}=3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{MgSO_4}=1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{120}{300+40}\cdot100\%\approx35,3\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{2\cdot98}{300+40}\cdot100\%\approx57,65\%\end{matrix}\right.\)

\(a)Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_{\text{ 4}}\right)_3+3H_2O\\ n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=3n_{Fe_2O_3}=0,03\left(mol\right)\\m_{ddH_2SO_4}=\dfrac{0,03.98}{19,6\%}=15\left(g\right)\\ b)m_{ddsaupu}=1,6+15=16,6\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,01\left(mol\right)\\ C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,01.400}{16,6}.100=24,1\%\)

\(n_{Fe2O3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)

a) Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)

               1              3                   1                 3

             0,01         0,03                0,01

\(n_{H2SO4}=\dfrac{0,01.3}{1}=0,03\left(mol\right)\)

⇒ \(m_{H2SO4}=0,03.98=2,94\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{2,94.100}{19,6}=15\left(g\right)\)

b) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,03.1}{3}=0,01\left(mol\right)\)

⇒ \(m_{Fe2\left(SO4\right)3}=0,01.400=4\left(g\right)\)

\(m_{ddspu}=1,6+15=16,6\left(g\right)\)

\(C_{Fe2\left(SO4\right)3}=\dfrac{4.100}{16,6}=24,1\)⁰/₀

 Chúc bạn học tốt

$m_{dd\ sau\ pư} = 2,4 + 300 = 302,4(gam)$
$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
$n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = \dfrac{2,4}{160} = 0,015(mol)$

$C\%_{Fe_2(SO_4)_3} = \dfrac{0,015.400}{302,4}.100\% = 1,98\%$

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=196.40\%=78,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{78,4}{98}=0,8\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:       0,2             0,6              0,2

Ta có: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\) ⇒ Fe₂O₃ hết, H₂SO₄ dư

m_{dd sau pứ} = 32 + 196 = 228 (g)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,2.400.100\%}{228}=35,09\%\)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,8-0,6\right).98.100\%}{228}=8,596\%\)

\(n_{Fe_2O_3}=\frac{2,4}{160}=0,015\left(mol\right)\)

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

(mol)       0,015       0,045             0,015

\(\Rightarrow m_{Fe_2\left(SO_4\right)_3}=0,045\times400=18\left(g\right)\)

\(\Rightarrow\%Fe_2\left(SO_4\right)_3=\frac{18}{300}\times100=6\%\)

a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{HCl} = 3n_{Al} = 0,6(mol)$
$m_{HCl} = 0,6.36,5 = 21,9(gam)$

b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

c)

$m_{dd\ sau\ pư} = 5,4 + 200 - 0,3.2 = 204,8(gam)$

$m_{HCl\ dư} = 200.20\% - 21,9 = 18,1(gam)$

$C\%_{HCl} = \dfrac{18,1}{204,8}.100\% = 8,84\%$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{204,8}.100\% = 13,04\%$

a) nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
VH2=0,3.22,4=6,72(lít)VH2=0,3.22,4=6,72(lít)

c)

mdd sau pư=5,4+200−0,3.2=204,8(gam)

mHCl dư=200.20%−21,9=18,1(gam)mHCl dư=200.20%−21,9=18,1(gam)

C%AlCl3=0,2.133,5204,8.100%=13,04%

\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

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

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{NaOH}=0,2\cdot40=8g\)

\(m_{ddNaOH}=4,6+0,2\cdot18-0,1\cdot2=8g\)

\(\Rightarrow C\%=\dfrac{m_{NaOH}}{m_{ddNaOH}}\cdot100\%=\dfrac{8}{8}\cdot100\%=100\%???\)

Sửa đề: Tính nồng độ mol của dung dịch NaOH???

\(C_{M_{NaOH}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}M\)

vì đề không cho \(D_{NaOH}=???\) nên không tính được dung dịch NaOH