PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) HCl còn dư, Magie p/ứ hết

\(\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Mg}+m_{ddHCl}-m_{H_2}=111,7\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{111,7}\cdot100\%\approx3,27\%\end{matrix}\right.\)

\(a.Mg+2HCl->MgCl_2+H_2\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1mol\\ n_{HCl}=\dfrac{109,5.10\%}{36,5}=0,3mol\\ n_{Mg}:1< n_{HCl}:2\\ Mg:hết\\ m_{ddsau}=2,4+109,5-2.0,1=111,7g\\ C\%_{HCl\left(dư\right)}=\dfrac{36,5.0,1}{111,7}.100\%=3,27\%\\ C\%_{MgCl_2}=\dfrac{95.0,1}{111,7}.100\%=8,50\%\)

\(a/\\MgO+2HCl \to MgCl_2+H_2O\\ n_{MgO}=\frac{8}{40}=0,2(mol)\\ b/\\ n_{HCl}=0,2.2=0,4(mol)\\ CM_{HCl}=\frac{0,4}{0,2}=2M\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

             1             2             1             1

           0,2            0,4

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

200ml =0,2l

\(C_{M_{ddHCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

 Chúc bạn học tốt

`a)PTPƯ: Zn + 2HCl -> ZnCl_2 + H_2↑`

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`b) n_[Zn] = 13 / 65 = 0,2 (mol)`

Theo `PTPƯ` có: `n_[HCl] = 2n_[Zn] = 2 . 0,2 = 0,4 (mol)`

  `-> m_[dd HCl] = [ 0,4 . 36,5 ] / [ 7,3 ] . 100 = 200 (g)`

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`c)` Theo `PTPƯ` có: `n_[H_2] = n_[ZnCl_2] = n_[Zn] = 0,2 (mol)`

`-> C%_[ZnCl_2] = [ 0,2 . 136 ] / [ 13 + 200 - 0,2 . 2 ] . 100 ~~ 12,79 %`

1a) \(n_{CO_2}=0,14\left(mol\right);n_{Ca\left(OH\right)_2}=0,08\left(mol\right)\Rightarrow n_{OH^-}=0,16\left(mol\right)\)

\(\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0,16}{0,14}=1,14\) => Tạo 2 muối

\(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

b) Gọi x, y là số mol Ca(HCO3)2 và CaCO3

\(\left\{{}\begin{matrix}x+y=0,08\\2x+y=0,14\end{matrix}\right.\)

=> x=0,06 ; y=0,02

\(m_{CaCO_3}=0,02.100=2\left(g\right)\)

c) \(CM_{Ca\left(HCO_3\right)_2}=\dfrac{0,06}{0,8}=0,075M\)

2. a) Bảo toàn nguyên tố C : \(n_{CO_2}=n_{Na_2CO_3}=0,25\left(mol\right)\)

\(n_{KOH}=0,5\left(mol\right)\)

Lập T = \(\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{0,5}{0,25}=2\) => Tạo 1 muối K2CO3, các chất phản ứng hết

b) \(m_{K_2CO_3}=0,25.138=34,5\left(g\right)\)

Câu 3 : 

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

100ml = 0,1l

\(n_{HCl}=2,5.0,1=0,25\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

             1          2              1           1

             0,1      0,25          0,1

b) Lập tỉ số so sánh : \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) 

                    ⇒ Mg phản ứng hết , HCl dư

                   ⇒ Tính toán dựa vào số mol của Mg

\(n_{HCl\left(dư\right)}=0,25-\left(0,1.2\right)=0,05\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,05.36,5=1,825\left(g\right)\)

c) \(n_{MgCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{MgCl2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

 Chúc bạn học tốt

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$

$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$

$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$

c) $Fe + CuSO_4 \to FeSO_4 + Cu$

$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$

$= 3,6 + 0,15.64 = 13,2(gam)$

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,2____0,4____0,2 (mol)

b, m_ZnCl2 = 0,2.136 = 27,2 (g)

c, \(V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

Bạn tham khảo nhé!