Bài 2:

a) PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

b) Dung dịch A là dung dịch bazơ

Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\) \(\Rightarrow n_{NaOH}=0,1\left(mol\right)\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{1}=0,1\left(M\right)\)

c) Sửa đề: dd H₂SO₄ 9,8%

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PTHH: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{9,8\%}=50\left(g\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{50}{1,14}\approx43,86\left(ml\right)\)

Bài 1:

PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot19,6\%}{98}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư

\(\Rightarrow n_{CuSO_4}=0,2\left(mol\right)=n_{H_2SO_4\left(dư\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,2\cdot160}{200+16}\cdot100\%\approx14,81\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2\cdot98}{200+16}\cdot100\%\approx9,07\%\end{matrix}\right.\)

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)  

             1            1                1            1

           0,2          0,2             0,2

b) \(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{19,6.100}{200}=9,8\)⁰/₀

c) \(n_{CuSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CuSO4}=0,2.160=32\left(g\right)\)

\(m_{ddspu}=16+200=216\left(g\right)\)

\(C_{CuSO4}=\dfrac{32.100}{216}=14,81\)⁰/₀

\(n_{CuO}=\frac{16}{80}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=\frac{200.19,6}{100}=39,2\left(g\right)\) => \(n_{H_2SO_4}=\frac{39,2}{98}=0,4\left(mol\right)\)

PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

_______0,2------>0,2---------->0,2____________(mol)

=> \(\left\{{}\begin{matrix}m_{CuSO_4}=0,2.160=32\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,4-0,2\right).98=19,6\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%\left(CuSO_4\right)=\frac{32}{16+200}.100\%=14,8\%\\C\%\left(H_2SO_4\right)=\frac{19,6}{16+200}.100\%=9,07\%\end{matrix}\right.\)

\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.4......................0.4\)

\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)

\(b.\)

\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)

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

\(0.16..........0.16..............0.16\)

\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)

\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)

a)

$SO_3 + H_2O \to H_2SO_4$

n SO3 = n H2SO4 = 200.19,6%/98 = 0,4(mol)

=> m = 0,4.80 = 32(gam)

b)

$MgO + H_2SO_4 \to MgSO_4 + H_2O$

n MgSO4 = n MgO = n H2SO4 = 80.19,6%/98 = 0,16(mol)

=> m MgO = 0,16.40 = 6,4(gam)

Sau pư, m dd =  6,4 + 80 = 86,4(gam)

=> C% MgSO4 = 0,16.120/86,4   .100% = 22,22%

a) 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

b) \(n_{NaOH}=\dfrac{100.20}{100.40}=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

______0,5----->0,25------>0,25

=> m_H2SO4 = 0,25.98 = 24,5 (g)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) m_Na2SO4 = 0,25.142 = 35,5 (g)

m_{dd sau pư} = 100 + 125 = 225 (g)

=> \(C\%\left(Na_2SO_4\right)=\dfrac{35,5}{225}.100\%=15,778\%\)

n_Fe = 11.2/56 = 0.2 (mol) 

Fe + H2SO4 => FeSO4 + H2 

0.2____0.2_______0.2___0.2 

mH2SO4 = 0.2*98 = 19.6 (g) 

mdd H2SO4 = 19.6*100/10 = 196 (g) 

m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g) 

mFeSO4 = 0.2*152 = 30.4 (g) 

C% FeSO4 = 30.4/206.8 * 100% = 14.7% 

Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml) 

CM FeSO4 = 0.2 / 0.1719 = 1.16 M 

n_Fe = 11.2/56 = 0.2 (mol) 

Fe + H2SO4 => FeSO4 + H2 

0.2____0.2_______0.2___0.2 

mH2SO4 = 0.2*98 = 19.6 (g) 

mdd H2SO4 = 19.6*100/10 = 196 (g) 

m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g) 

mFeSO4 = 0.2*152 = 30.4 (g) 

C% FeSO4 = 30.4/206.8 * 100% = 14.7% 

Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml) 

CM FeSO4 = 0.2 / 0.1719 = 1.16 M 

Fe+H2SO4->FeSO4+H2

0,1----0,1-------0,1-----0,1 

n Fe=5,6\56=0,1 mol

=>VH2=0,1.22,4=2,24l

=>m H2SO4=0,1.98=9,8g

=>C%=9,8\100 .100=9,8%

\(n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\)

PTHH:

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,125     0,375             0,125           0,375 

\(m_{ddH_2SO_4}=\dfrac{0,375.98.100}{25}=147\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,125.406}{20+147}\approx30,39\%\)