a) PTHH: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\uparrow\)

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)=n_{SO_2}\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)

b) Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=0,1\left(mol\right)\\n_{Ca\left(OH\right)_2}=1,4\cdot0,1=0,14\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa

PTHH: \(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3\downarrow+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CaSO_3}=0,1\left(mol\right)=n_{Ns_2SO_4}\\n_{Ca\left(OH\right)_2\left(dư\right)}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_3}=0,1\cdot120=12\left(g\right)\\m_{Na_2SO_4}=0,1\cdot142=14,2\left(g\right)\\m_{Ca\left(OH\right)_2\left(dư\right)}=0,04\cdot74=2,96\left(g\right)\end{matrix}\right.\)

\(n_{Na_2SO_3}=0,1\left(mol\right)\\ PTHH:Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\)

(mol)          0,1                              0,1             0,1       0,1

\(a.V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

\(b.n_{Ca\left(OH\right)_2}=0,14\left(mol\right)\)

Do \(\dfrac{n_{OH}}{n_{SO_2}}=\dfrac{0,28}{0,1}=2.8>2\rightarrow\) Tạo muối trung hòa và Ca(OH)₂ dư 0,04(mol)

\(PTHH:Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)

(mol)           0,1           0,1         0,1            0,1

\(m_{Ca\left(OH\right)_2\left(du\right)}=0,04.74=2,96\left(g\right)\\ m_{CaSO_3}=12\left(g\right)\\ m_{H_2O}=0,1.18=1,8\left(g\right)\)

Ta có ptpu

MgCO3+ 2HCl ----> MgCl2 + H2O+ CO2

\(n_{MgCO3}\)= \(\frac{9,6}{84}\)= \(\frac{0,8}{7}\) ( mol)

\(m_{HCl}\)= \(\frac{14,6}{100}.100\)= 14,6(g)

=> \(n_{HCl}\)= \(\frac{14,6}{36,5}=0,4\left(mol\right)\)

 Theopt ta thấy sau phản ứng HCl dư và dư \(\frac{1,2}{7}\) mol==> m dư= 6,26 (g)

=> \(n_{CO2}\)= \(n_{MgCO3}\)= \(\frac{0,8}{7}\) mol

=> \(V_{CO2}\)= \(\frac{0,8}{7}.22,4=2,56\left(l\right)\)

b)

Ta có \(m_{CO2}\)= \(\frac{0,8}{7}.44=\frac{35,2}{7}\left(g\right)\)

\(m_{H2O}\)= \(\frac{0,8}{7}.18=\frac{14,4}{7}\)( g)

\(m_{MgCl2}\)= \(\frac{0,8}{7}.95=\frac{76}{7}\)(g)

=> \(m_{dd_{MgCl2}}\)= (9,6+100)-( \(\frac{49,6}{7}\))= 102,5(g)

=> \(C\%_{MgCl2}\)= \(\frac{\frac{76}{7}}{102,5}\). 100%= 10,6 ( %)

\(C\%_{HCl_{dư}}\)= \(\frac{6,26}{102,5}.100\)=6,107 ( %)

đề có hơi vô lý một tý nha: hỗn hợp thì phải 2 chất trở lên, nhưng trong đề chỉ có mỗi MgCO3....Đề ghi vậy thì anh làm theo đề nha

mHCl=(14.6*100)/100=14.6g=>nHCl=0.4 mol

nMgCO3=9.6/84=4/35\(\approx\) 0.114mol

theo pthh,HCl dư 2/7 mol=>mHCl dư=73/7g

MgCO3+2HCl --> MgCl2 + CO2+H2O

  4/35        0.4              4/35   4/35    

a)Vco2=(4/35)*22.4=2.56l

b)C%HCl dư=(73/7)/100*100=76/7\(\approx\) 10.42%

mMgCl2=4/35*95=76.7g

C%MgCl2=(76.7)/100*100=76/7\(\approx\) 10.85%

Chúc em học tốt!!1

\(a,n_{AlCl_3}=1\cdot0,2=0,2\left(mol\right)\\ m_{Na_2CO_3}=\dfrac{200\cdot6\%}{100\%}=12\left(g\right)\\ \Rightarrow n_{Na_2CO_3}=\dfrac{12}{106}\approx0,1\left(mol\right)\\ PTHH:3AlCl_3+2Na_2CO_3+H_2O\rightarrow2Al\left(OH\right)_3\downarrow+6NaCl+3CO_2\uparrow\)

Vì \(\dfrac{n_{AlCl_3}}{3}>\dfrac{n_{Na_2CO_3}}{2}\) nên sau phản ứng \(AlCl_3\) dư

\(\Rightarrow n_{Al\left(OH\right)_3}=n_{Na_2CO_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al\left(OH\right)_3}=0,1\cdot78=7,8\left(g\right)\\ b,n_{NaCl}=3n_{Na_2CO_3}=0,3\left(mol\right)\\ \Rightarrow m_{NaCl}=0,3\cdot58,5=17,55\left(g\right)\)

Mấy bạn giúp mình đi :(

CuSO4+Ca(OH)2--> Cu(OH)2+ CaSO4 (1)

a) nCuSO4=0,5(mol)

nCa(OH)2=0,1(mol)

lập tỉ lệ :

\(\dfrac{0,5}{1}>\dfrac{0,1}{1}\)

=>CuSO4 dư , Ca(OH)2 hết => bài toán tính theo Ca(OH)2

theo (1) : nCuSO4(pư)=nCu(OH)2=nCaSO4=nCa(OH)2=0,1(mol)

=>mCu(OH)2=9,8(g)

mCaSO4=13,6(g)

b)Vdd=500ml=0,5l

nCuSO4(dư)=0,4(mol)

=> C_{M dd CuSO4 dư}=\(\dfrac{0,4}{0,5}=0,8\left(M\right)\)

a) \(n_{CH_3COOH}=\dfrac{100.12\%}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaHCO₃ --> CH₃COONa + CO₂ + H₂O

                 0,2-------->0,2----------->0,2--------->0,2

=> m_NaHCO3 = 0,2.84 = 16,8 (g)

=> \(m_{dd.NaHCO_3}=\dfrac{16,8.100}{8,4}=200\left(g\right)\)

b) m_CH3COONa = 0,2.82 = 16,4 (g)

m_{dd sau pư} = 100 + 200 - 0,2.44 = 291,2 (g)

=> \(C\%_{dd.muối}=\dfrac{16,4}{291,2}.100\%=5,632\%\)

\(m_{ct}=\dfrac{12.100}{100}=12\left(g\right)\)

\(n_{CH3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)

Pt : \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O|\)

                 1                   1                       1                1           1

               0,2                 0,2                    0,2              0,2

a) \(n_{NaHCO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{NaHCO3}=0,2.84=16,8\left(g\right)\)

\(m_{ddNaHCO3}=\dfrac{16,8.100}{8,4}=200\left(g\right)\)

b) \(n_{CH3COONa}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{CH3COONa}=0,2.82=16,4\left(g\right)\)

\(m_{ddspu}=100+200-\left(0,2.44\right)=291,2\left(g\right)\)

\(C_{CH3COONa}=\dfrac{16,4.100}{291,2}=5,63\)⁰/₀

 Chúc bạn học tốt

\(a,Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{Fe}=n_{H_2}=0,2\left(mol\right)\\ \%m_{Fe}=\dfrac{0,2.56}{12,8}.100\%=87,5\%\\ \%m_{Fe_2O_3}=100\%-87,5\%=12,5\%\\ c,n_{Fe_2O_3}=\dfrac{12,8-11,2}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=n_{Fe}+3n_{Fe_2O_3}=0,2+3.0,01=0,23\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,23}{0,46}=0,5\left(M\right)\)