a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)

PTHH: Mg + 2CH₃COOH --> (CH₃COO)₂Mg + H₂

                            0,02<------------0,01----->0,01

=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,02}{0,05}=0,4M\)

b) V_H2 = 0,01.22,4 = 0,224 (l)

\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)

\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)

Zn +H2SO4--->ZnSO4+H2(1)

Zn+2HCl----.ZnCl2 +H2(2)

a) Ta co

n\(_{Zn}\frac{3,25}{65}=0,05\left(mol\right)\)

Theo pthh 1,2

n\(_{H2}=n_{Zn}=0,05\left(mol\right)\)

V\(_{H2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

b) Gọi n\(_{Zn}pt1=x\)

Gọi n\(_{Zn}pt2=y\)

=>x+y=0,05(*)

Mặt khác

m\(_{ZnSO4}=161x\)

m\(_{ZnCl2}=136y\)

=> 161x+136y=7,55(**)

Từ 1 và 2 ta có hệ pt

\(\left\{{}\begin{matrix}x+y=0,05\\161x+136y=7,55\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,03\\y=0,02\end{matrix}\right.\)

Theo pthh1

n\(_{H2SO4}=n_{Zn}=0,03\left(mol\right)\)

=>C\(_{M\left(H2SO4\right)}=\frac{0,03}{0,5}=0,06\left(M\right)\)

Theo pthh2

n\(_{HCl}=2n_{Zn}=0,02\left(mol\right)\)

C\(_{M\left(HCl\right)}=\frac{0,02}{0,5}=0,04\left(M\right)\)

Chúc bạn học tốt

a) $Zn + 2HCl \to ZnCl_2 + H_2$

b) Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{16,25}{65} = 0,25(mol)$
$\Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)$

c) $n_{HCl} = 2n_{Zn} = 0,5(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,5}{0,5} = 1M$

2CH3COOH + Mg => (CH3COO)₂Mg + H2

m_muối = 0.71 (g) => n_muối = m/M = 0.71/142 = 0.005 (mol)

Theo pthh ==> n_CH3COOH = 0.01 (mol)

C_M = n/V = 0.01/0.05 = 0.2 M

CH3COOH + NaOH => CH3COONa + H2O

Trong 50 ml axit trên có 0.01 mol

Theo pt ===> n_NaOH = 0.01 (mol)

V_{dd NaOH} = n/C_M = 0.01/0.75 = 1/75 (l)

Cái này xét nhiều TH lắm như ra muối Na₃PO₄ , Na₂HPO₄,NaH₂PO₄

\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

Na₂CO₃  +  2HCl  →  2NaCl  +  H₂O  + CO₂

a) nNa₂CO₃ = 0,2.0,5 = 0,1 mol 

Theo tỉ lệ phản ứng => nHCl = 2nNa₂CO₃ = 0,1.2 = 0,2 mol

=> C_HCl = \(\dfrac{n}{V}\) = \(\dfrac{0,2}{0,2}\)=  1M

b) nNaCl = 2Na₂CO₃ = 0,2 mol

=> mNaCl = 0,2.58,5 = 11,7 gam

c) Thể tích dung dịch sau phản ứng = 0,2 + 0,2 = 0,4 lít

=> C_NaCl = \(\dfrac{0,2}{0,4}\)= 0,5M