CH3COOH+NaHCO3->CH3COONa+H2O+CO2

0,1----------------0,1-------------------0,1--------------0,1

n CO2=0,1 mol

=>C% axit=\(\dfrac{0,1.60}{200}.100\)=3%

m NaHCO3=0,1.84=8,4g

c) C% CH3COONa=\(\dfrac{0,1.82}{200+8,4}.100=3,934\%\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(a) n_{CH_3COOH} = \dfrac{200.12\%}{60} = 0,4(mol)\\ 2CH_3COOH + CaCO_3 \to (CH_3COO)_2Ca + CO_2 + H_2O\\ n_{CaCO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ \Rightarrow a = \dfrac{0,2.100}{100\%-20\%} =25(gam)\\ V_B = 0,2.22,4 = 4,48(lít)\\ b) m_{dd\ sau\ pư} = m_{CaCO_3} + m_{dd\ CH_3COOH} - m_{CO_2} = 0,2.100 + 200 - 0,2.2 = 219,6(gam)\\ C\%_{(CH_3COO)_2Ca} = \dfrac{0,2.158}{219,6}.100\% = 36\%\)

nZn = 3,25/65=0,05 mol

2Zn + 2CH3COOH --> 2CH3COOZn + H2

0,05      0,05                   0,05               0,025            mol

=> VH2= 0,025*22,4=0,56 lít

mdd=(0,05*60*100)/20=15 g

b)mCH3COOZn = 0,05*124=6,2 g

Zn+2HCl→ZnCl2+H2

 2Al+6HCl→2AlCl3+3H2

Gọi số mol HCllà x

n H2=\(\dfrac{1}{2}\)n HCl=0,5x

Bảo toàn khối lượng: mkl+mHCl=mmuối+mH2

6,05+36,5x=13,15+0,5x.2

→x=0,2 mol

mHCl=0,2.36,5=7,3 gam

=>m =mddHCl=\(\dfrac{7,3}{10\%}\)=73gam

a)

$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$

b)

n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)

m dd NaHCO3 = 0,2.84/8% = 210(gam)

c)

n CO2 = n CH3COOH = 0,2(mol)

=> V CO2 = 0,2.22,4 = 4,48(lít)

d)

m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100  + 210 - 0,2.44 = 301,2(gam)

C% CH3COONa = 0,2.82/301,2   .100% = 5,44%

Bài 3 : 

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)

        2            3                   1               3

       0,1       0,15                 0,05       0,15

a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

\(m_{H2}=0,15.2=0,3\left(g\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

⇒ \(m=0,15.98=14,7\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)⁰/₀

c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)

\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)⁰/₀

 Chúc bạn học tốt

\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)