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

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

      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

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

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

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

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)

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!

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)

\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

 0,4           0,4         0,4        0,4

a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\) 

 \(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)

b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

   \(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)

   \(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)

c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

     \(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)

         0,4          0,3         0,3              0,3

     \(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

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

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

PTHH :

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

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

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

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

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

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

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

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

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