PT \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+H_2O+CO_2\uparrow\)

a, \(m_{CH_3COOH}=\frac{6\times150}{100}=9\left(g\right)\)\(\Rightarrow n_{CH_3COOH}=\frac{9}{60}=0,15\left(mol\right)\)

Theo PT \(n_{NaHCO_3}=n_{CH_3COOH}=0,15\left(mol\right)\Rightarrow m_{NaHCO_3}=0,15\times84=12,6\left(g\right)\)\(\Rightarrow m_{ddNaHCO_3}=\frac{12,6\times100}{8,4}=150\left(g\right)\)

b, Theo PT \(n_{CH_3COONa}=n_{CH_3COOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{CH_3COONa}=0,15\times82=12,3\left(g\right)\)

Có \(m_{ddsaupu}=150+150=300\left(g\right)\)

\(\Rightarrow C\%=\frac{12,3}{300}\times100=4,1\%\)

mCH3COOH= 150*6/100=9g

nCH3COOH= 9/60=0.15 mol

CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O

0.15_________0.15__________0.15______0.15

mNaHCO3= 0.15*84=12.6g

mdd NaHCO3= 12.6*100/8.4=150g

m dung dịch sau phản ứng=mdd CH3COOH + mdd NaHCO3 - mCO2= 150+150-0.15*44==293.4g

C%CH3COONa= 12.3/293.4*100%= 4.19%

CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O

mCH3COOH = 150 x 6/100 = 9 (g)

===> nCH3COOH = m/M = 9/60 = 0.15 (mol)

Theo phương trình ==> nNaHCO3 = 0.15 (mol)

mNaHCO3 = n.M = 0.15 x 84 = 12.6 (g)

===> mddNaHCO3 = 12.6 x 100/8.4 = 150 (g)

mdd sau pứ = 150 + 150 - 0.3 = 299.7 (g)

mCH3COONa = n.M = 0.15 x 82 = 12.3 (g)

C%ddCH3COONa = 4.104 %

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%

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

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

PTHH :

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

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

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

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

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

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

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

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

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

CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O

m_CH3COOH = 100x12/100 = 12 (g)

==> n_CH3COOH = m/M = 12/60 = 0.2 (mol)

Theo pt: => n_NaHCO3 = 0.2 (mol)

==> m_NaHCO3 = n.M = 0.2x84 =16.8 (g)

==> m_{dd NaHCO3} = 16.8x100/8.4 = 200 (g)

Ta có: n_CH3COONa = 0.2 (mol)

==> m_CH3COONa = n.M = 0.2 x 82 = 16.4 (g)

m_{dd sau pứ} = 200 + 100 - 0.2 x 44 =291.2 (g)

C% = 16.4 x 100/ 291.2 = 5.63%

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

PTHH: \(CH_3COOH+KHCO_3\rightarrow CH_3COOK+H_2O+CO_2\uparrow\)

a) Ta có: \(n_{CH_3COOH}=\dfrac{200\cdot24\%}{60}=0,8\left(mol\right)=n_{KHCO_3}\)

\(\Rightarrow m_{ddKHCO_3}=\dfrac{0,8\cdot100}{16,8\%}\approx476.2\left(g\right)\)

b) Theo PTHH: \(n_{CH_3COOK}=0,8\left(mol\right)=n_{CO_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COOK}=0,8\cdot98=78,4\left(g\right)\\m_{CO_2}=0,8\cdot44=35,2\left(g\right)\end{matrix}\right.\)

 Mặt khác: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)

\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)