\(m_{CH_3COOH}=\dfrac{80.9}{100}=7,2\left(g\right)\)

\(n_{CH_3COOH}=\dfrac{7,2}{60}=0,12\left(mol\right)\)

PTHH :

\(15CH_3COOH+10NaHCO_3\rightarrow10CH_3COONa+2H_2O+20CO_2\uparrow\)

0,12                        0,08                 0,08                     0,016     0,16 

\(a,m_{NaHCO_3}=84.0,08=6,72\left(g\right)\)

\(m_{ddNaHCO_3}=\dfrac{6,72.100}{4,2}=160\left(g\right)\)

\(b,m_{CH_3COONa}=0,08.82=6,56\left(g\right)\)

\(m_{H_2O}=0,016.18=0,288\left(g\right)\)

\(m_{CO_2}=0,16.44=7,04\left(g\right)\)

\(m_{ddCH_3COONa}=80+160-0,288-7,04=232,672\left(g\right)\)

\(C\%=\dfrac{6,56}{232,672}\approx2,82\%\)

\(n_{CH_3COOH}=\dfrac{50.12\%}{60}=0,1mol\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

      0,1                  0,05                                                  ( mol )

\(m_{Na_2CO_3}=0,05.106=5,3g\)

\(m_{dd_{Na_2CO_3}}=\dfrac{5,3}{8,4\%}=63,09g\)

\(m_{CH_3COOH}=\dfrac{12.50}{100}=6\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ ---> 2CH₃COONa + CO₂ + H₂O

              0,1------------>0,05

=> \(m_{ddNa_2CO_3}=\dfrac{0,05.106}{8,4\%}=63,1\left(g\right)\)

\(NaHCO_3 + HCl \to NaCl + CO_2 + H_2O\\ n_{CO_2 } = n_{NaHCO_3} = \dfrac{252}{84} = 3(mol)\\ \Rightarrow V_{CO_2} = 3.22,4 = 67,2(lít)\)

nNaHCO3 = 252/84 = 3 (mol) 

NaHCO3 + HCl => NaCl + CO2 + H2O 

3______________________3

VCO2 = 3*22.4 = 67.2 (l) 

PT: \(NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2\)

Ta có: \(n_{NaHCO_3}=\dfrac{252}{84}=3\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{NaHCO_3}=3\left(mol\right)\)

\(\Rightarrow V_{CO_2}=3.22,4=67,2\left(l\right)\)

Bạn tham khảo nhé!

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%

a)Na2CO3+2HCl--->2NaCl+H2O+CO2

x------------------------------------------------x-

CaCO3+2HCl--->CaCl2+H2O+CO2

y-----------------------------------------y

Ta có n CO2=6,72/22,4=0,3(mol)

Theo bài ra ta có hpt

\(\left\{{}\begin{matrix}106x+100y=30,6\\x+y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

%m Na2CO3=0,1.106/30,6.100%=34,64%

%m CaCO3=100%-34,64%=65,36%

b) n HCl=2n CO2=0,6(mol)

m HCl=0,6.36,5=21,9(g)

m dd HCl=21,9.100/20=109,5(g)

m dd sau pư=m hh+m dd HCl-m CO2

=30,6+109,5-18=122,1(g)

%m NaCl=0,2.58,5/122,1.100%=9,58%

%m CaCl2=0,2.111/122,1.100%=18,18%

n

giup minh voi minh cam on a

10)

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

x_____________2x______2x________x__x(mol)

CaCO3 + 2 HCl -> CaCl2 +H2O + CO2

y_________2y______y_y_________y(mol)

nCO2= x+y=6,72/22,4=0,3 (1)

mNa2CO3 + mCaCO3= mhh

<=> 106x+100y= 30,6 (2)

từ (1)., (2) có hpt:

\(\left\{{}\begin{matrix}x+y=0,3\\106x+100y=30,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> mNa2CO3= 0,1. 106 =10,6(g)

=> \(\%mNa2CO3=\frac{10,6}{30,6}.100\approx34,641\%\\ \Rightarrow\%mCaCO3\approx100\%-34,641\%\approx65,359\%\)

b) nHCl= 2x+2y= 2.0,1+2.0,2=0,6(mol)

=> mHCl= 0,6.36,5= 21,9(g)

=> mddHCl= (21,9.100)/20= 109,5(g)

=> mddsau= m(hh)+ mddHCl - mCO2= 30,6+109,5 - 0,3.44= 126,9(g)

mNaCl= 2x.58,5= 2.0,1.58,5= 11,7(g)

mCaCl2= 2y.111= 2.0,2.111= 44,4(g)

=> \(C\%ddNaCl=\frac{11,7}{126,9}.100\approx9,22\%\\ C\%ddCaCl2=\frac{44,4}{126,9}.100\approx34,988\%\)

a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

b. Đổi 100ml = 0,1 lít

Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

Đề này có cho D của dung dịch ko em?

Sửa đề : 200ml thành 200g 

a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O\) (2)

b) 1/2 lượng khí B: \(n_{H_2\left(2\right)}=3n_{Fe_2O_3}=3.\dfrac{38,4}{160}=0,72\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=0,72.2=1,44\left(mol\right)\)

\(n_{H_2SO_4}=n_{H_2\left(2\right)}=1,44\left(mol\right)\)

=> \(C\%H_2SO_4=\dfrac{1,44.98}{200}.100=70,56\%\)

\(n_{Al}=\dfrac{2}{3}n_{H_2\left(2\right)}=0,96\left(mol\right)\)

=> \(m_{Al}=0,96.27=25,92\left(g\right)\)

Có lẽ đề bài cho 200 g dd H₂SO₄ chứ không phải 200 ml bạn nhỉ?

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(FeO+H_2\underrightarrow{t^o}Fe+H_2O\) (2)

b, Ta có: \(n_{FeO}=\dfrac{38,4}{72}=\dfrac{8}{15}\left(mol\right)\)

Theo PT: \(n_{H_2\left(2\right)}=n_{FeO}=\dfrac{8}{15}\left(mol\right)\)

\(\Rightarrow n_{H_2\left(1\right)}=\dfrac{8}{15}.2=\dfrac{16}{15}\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{32}{45}\left(mol\right)\\n_{H_2SO_4}=n_{H_2}=\dfrac{16}{15}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x=m_{Al}=\dfrac{32}{45}.27=19,2\left(g\right)\)

\(m_{H_2SO_4}=\dfrac{16}{15}.98=\dfrac{1568}{15}\left(g\right)\)

\(\Rightarrow y=C\%_{H_2SO_4}=\dfrac{\dfrac{1568}{15}}{200}.100\%\approx52,27\%\)

Bạn tham khảo nhé!