a)

Gọi $n_{Na_2CO_3} = a(mol) \Rightarrow n_{K_2CO_3}= 2a(mol)$

$Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
$K_2CO_3 + 2HCl \to 2KCl + CO_2 + H_2O$

$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
Theo PTHH : 

$n_{Na_2CO_3} + n_{K_2CO_3} = n_{CO_2} = n_{CaCO_3} $
$\Rightarrow a + 2a = 0,3$
$\Rightarrow a = 0,1$
$\Rightarrow m_{hh} = 0,1.106 + 0,1.2.138 = 38,2(gam)$

b)

$n_{HCl} =2 n_{Na_2CO_3} + 2n_{K_2CO_3} = 0,6(mol)$
$V_{dd\ HCl} = \dfrac{0,6}{1,5} = 0,4(lít)$

a,Gọi \(n_{Na_2CO_3}=x\left(mol\right);n_{K_2CO_3}=2x\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

PTHH: K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

PTHH: CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

Theo PTHH ta có:

 \(n_{Na_2CO_3}+n_{K_2CO_3}=n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

⇒ x + 2x = 0,3

 ⇔ x = 0,1 (mol)

⇒ m_{hh muối} = 0,1.106 + 0,1.2.138 = 38,2 (g)

b, \(n_{HCl}=2n_{CO_2}=2n_{CaCO_3}=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)\)

Câu này em đã từng hỏi r ,em xem có nhầm lẫn gì ko nhé

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2mol\\ a.MgCO_3+H_2SO_4->MgSO_4+H_2O+CO_2\\ 2NaOH+H_2SO_{\text{4 }}->Na_2SO_4+2H_2O\\ b.n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.80.0,1:40=0,1mol\\ n_{H_2SO_4\left(MgCO_3\right)}=0,2mol\\ c.C\%=\dfrac{98.0,3}{200}.100\%=14,7\%\\ V=0,2.22,4=4,48L\\ d.m_{ddsau}=200+16,8-44.0,2+80=288g\\ C\%_{Na_2SO_4}=\dfrac{40.0,1}{288}.100\%=1,39\%\\ C\%_{MgSO_4}=\dfrac{120.0,2}{288}.100\%=8,33\%\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{80}{40}=2\left(mol\right)\)

PTHH :

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\)

0,2              0,2                                         0,2 

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

2                 1                 1

Vậy có 0,2 mol H2SO4 phản ứng với MgCO3 

       có 1 mol H2SO4 phản ứng với NaOH 

\(m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)

\(c,C\%_{H_2SO_4}=\dfrac{117,6}{200}.100\%=58,8\%\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(d,m_{Na_2SO_4}=1.142=142\left(g\right)\)

\(m_{ddNaOH}=\dfrac{80.100}{10}=800\left(g\right)\)

\(m_{ddH_2SO_4dư}=1.98:58,8\%\approx166,67\left(g\right)\)

\(m_{ddNa_2SO_4}=800+166,67=966,67\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{142}{966,67}.100\%\approx14,69\%\)

1.n_CO2=0,1 (mol ) 

TH1: Số mol của CO₂ dư => Khối lượng muối khan tối đa tạo được là:

m_muối=0,1.84=8,4<9,5 (loại ) 

TH2: CO₂ hết 

Gọi số mol CO₂ tạo muối Na2CO3;NaHCO3 lần lượt là x, y

2NaOH+CO2→Na2CO3+H2O

NaOH+CO2→NaHCO3

Ta có : \(\left\{{}\begin{matrix}x+y=0,1\\106x+84y=9,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

⇒n_NaOH=2.0,05+0,05=0,15 (mol)

⇒CM_NaOH=\(\dfrac{0,15}{0,1}\)=1,5M

Câu 2 thật ra anh thấy chưa chặt chẽ nha!

2. Vì CO2 bị hấp thụ hoàn toàn

=> CO2 hết, NaOH dư

nCO2 = 0,1 (mol)

CO2 + 2NaOH -------> Na2CO3 + H2O

\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)

=> \(m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(1,\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2,\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ 3,\\ m_{MgCl_2}=95.0,25=23,75\left(g\right)\\ 4,\\ H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu}=n_{H_2}=0,25\left(g\right)\\ m_{Cu}=0,25.64=16\left(g\right)\)

1. \(Mg+2HCl\rightarrow MgCl_2+H_2\)

2. \(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{18,25}{36,5}\approx0,5\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}\)

                \(\Rightarrow n_{H_2}=\dfrac{1}{2}.0,5=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,25.22,4=5,6\left(l\right)\)

3. Theo PTHH: \(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}\)

                     \(\Rightarrow n_{MgCl_2}=0,25\left(mol\right)\)

\(m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,25.95=23,75\left(g\right)\)

4. \(H_2+CuO\rightarrow Cu+H_2O\)

Theo PTHH: \(n_{Cu}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,25.64=16\left(g\right)\)

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

     0,3      0,6             0,3        0,3

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

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