Bài 1:

Ta có: \(n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\)

Bảo toàn Hidro: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow m_{HCl\left(p.ứ\right)}=0,4\cdot36,5=14,6\left(g\right)\)

Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl\left(p.ứ\right)}-m_{H_2}=24,7\left(g\right)\)

Bài 2:

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1\cdot56}{12,8}\cdot100\%=43,75\%\) \(\Rightarrow\%m_{FeO}=56,25\%\)

Bài 3:

PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

Ta có: \(n_{KOH}=0,1\cdot1=0,1\left(mol\right)=n_{HCl}\)

*Bạn bổ sung thêm thể tích ddHCl thêm nhé !! 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{HCl}=0,5.1=0,5\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\\ \Rightarrow HCldư\\ \Rightarrow n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,5-2.0,2=0,1\left(mol\right)\\ a.V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.m_{ddsau}=11,2+500.1,132-0,2.2=576,8\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{576,8}.100\approx0,633\%\\ C\%_{ddFeCl_2}=\dfrac{0,2.127}{576,8}.100\approx4,404\%\)

a)

$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol) ; n_{HCl} = 0,5(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
Ta thấy : 

$n_{Fe} : 1 < n_{HCl} : 2$ nên HCl dư

$n_{H_2} = n_{Fe} = 0,2(mol) \Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)$

b)

$m_{dd\ HCl} = 500.1,132. = 566(gam)$
$m_{dd\ sau\ pư} = 11,2 + 566 - 0,2.2 = 576,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{576,8}.100\% = 4,4\%$
$C\%_{HCl} = \dfrac{(0,5 - 0,2.2).36,5}{576,8}.100\% = 0,632\%$

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

500ml =0,5l

\(n_{HCl}=1.0,5=0,5\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1         2             1           1

       0,2    0,5            0,2        0,2

a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\)

            ⇒ Fe phản ứng hết , HCl dư

           ⇒ Tính toán dựa vào số mol của Fe

\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{h2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

b) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,5-\left(0,2.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=1,132.500=566\left(g\right)\)

\(m_{ddspu}=11,2+566-\left(0,2.2\right)=576,8\left(g\right)\)

\(C_{FeCl2}=\dfrac{25,4.100}{576,8}=4,4\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{576,8}=0,63\)⁰/₀

 Chúc bạn học tốt

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Zn}:b\left(mol\right)\end{matrix}\right.\)

Giải hệ PT:

\(\left\{{}\begin{matrix}56a+65b=18,6\\127a+136b=39,9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\%m_{Fe}=\frac{0,1.56}{18,6}.100\%=30,11\%\)

\(\%m_{Fe}=100\%-30,11\%=69,89\%\)

b, \(n_{HCl}=0,1.2+0,2.2=0,6\left(mol\right)\)

\(V_{dd\left(HCl\right)}=\frac{0,6}{4}=0,15\left(l\right)=150\left(ml\right)\)

\(m_{dd\left(HCl\right)}=150.1,1=165\left(g\right)\)

c,\(ZnCl_2+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2AgCl\)

\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_3+2AgCl\)

\(n_{AgCl}=0,1.2+0,2.2=0,6\left(mol\right)\)

\(\Rightarrow m_{AgCl}=0,6.143,5=86,1\left(g\right)\)

a, Ta có: \(n_{CO}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{CO_2}\)

Theo ĐLBT KL, có: m_hh + m_CO = m_Fe + m_CO2

⇒ m_Fe = 18,2 + 0,3.28 - 0,3.44 = 13,4 (g)

b, Giả sử: \(\left\{{}\begin{matrix}n_{Ca}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

⇒ x + y = 0,2 (1)

PT: \(Ca+2HCl\rightarrow CaCl_2+H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(CaCl_2+Na_2CO_3\rightarrow CaCO_{3\downarrow}+2NaCl\)

\(MgCl_2+Na_2CO_3\rightarrow MgCO_{3\downarrow}+2NaCl\)

Theo PT: \(\left\{{}\begin{matrix}n_{CaCO_3}=n_{Ca}=x\left(mol\right)\\n_{MgCO_3}=n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

⇒ 100x + 84y = 18,4 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{CaCl_2}=n_{Ca}=0,1\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ a = m_CaCl2 + m_MgCl2 = 0,1.111 + 0,1.95 = 20,6 (g)

Bạn tham khảo nhé!

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

a, \(m_{FeCl_2}=0,1.\left(56+35,5.2\right)=12,7\left(g\right)\)

b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

giúp e với ạ

a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Theo PTHH ta có: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,1.95=9,5\left(g\right)\)

b) Theo PTHH ta có: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)

c) \(2H_2+O_2\rightarrow2H_2O\)

Theo PTHH: \(n_{H_2O}=\dfrac{0,1.2}{2}=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,1.18=1,8\left(g\right)\)

a)mMg= 2,4/24=0,1(mol) nMgCl2=0,1.1/1=0,1 (mol) mMgCl2= 0,1.95=9,5(g) b)nH2=0,1.1/1=0,1(mol) v=n.22,4=2,24(lít

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