1

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

a

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

0,3-->0,6---->0,3------->0,3

b

\(C\%_{dd.HCl}=\dfrac{0,6.36,5.100\%}{400}=5,475\%\)

c

\(m_{dd}=16,8+400-0,3.2=416,2\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,3.127.100\%}{416,2}=9,15\%\)

2

\(n_{HCl}=\dfrac{200.7,3\%}{100\%}:36,5=0,4\left(mol\right)\)

a

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

0,2<--0,4------>0,2------>0,2

b

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c

\(x=m_{Mg}=0,2.24=4,8\left(g\right)\)

d

\(m_{dd}=4,8+200-0,2.2=204,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)

3

\(n_{H_2SO_4}=\dfrac{300.9,8\%}{100\%}:98=0,3\left(mol\right)\)

a

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2<--0,3--------->0,1------------>0,3

b

\(m_{Al}=0,2.27=5,4\left(g\right)\)

c

\(m_{dd}=5,4+300-0,3.2=304,8\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342.100\%}{304,8}=11,22\%\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH :

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

0,25      0,5                   0,25

\(a,m_{Fe}=0,25.56=14\left(g\right)\)

\(b,C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

            1         2             1          1

          0,15     0,3                      0,15

b) \(n_{Fe}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe}=0,15.56=8,4\left(g\right)\)

c) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

50ml = 0,05l

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)

 Chúc bạn học tốt

Các bạn ơi giúp tôi dj mà đây là đề thj của trường tôi đó

Đây là hóa học mà

hsg hay đại trà

Bài 1:

Ta có: \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

BTNT H, có: \(n_{HCl}=2n_{H_2O}\Rightarrow n_{H_2O}=0,05\left(mol\right)\)

Theo ĐL BTKL, có: m _oxit + m_HCl = m_muối + m_H2O

⇒ m_muối = 2,8 + 0,1.36,5 - 0,05.18 = 5,55 (g)

Bài 2:

\(m_{KOH}=200.5,6\%=11,2\left(g\right)\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\)

Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

Bài 3:

Ta có: \(C\%_{NaCl}=\dfrac{m_{NaCl}}{m_{NaCl}+m_{H_2O}}.100\%=32\%\)

\(\Rightarrow\dfrac{m_{NaCl}}{m_{NaCl}+200}=0,32\)

\(\Rightarrow m_{NaCl}\approx94,12\left(g\right)\)

Bài 4:

Ta có: \(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

Theo PT: \(n_{CaO\left(LT\right)}=n_{CaCO_3}=0,15\left(mol\right)\)

\(\Rightarrow m_{CaO\left(LT\right)}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow H=\dfrac{6,72}{8,4}.100\%=80\%\)

\(Fe_2O_3 + 6HCl \rightarrow 2FeCl_3 + 3H_2O\)

\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)

\(NaOH + HCl \rightarrow NaCl + H_2O\)

\(n_{NaOH} = 0,2 . 1 = 0,2 mol\)

\(n_{HCl dư} = n_{NaOH}= 0,2 mol\)

\(\Rightarrow n_{HCl pư}= n_{HCl ban đầu} - n_{HCl dư}= 1,4- 0,2 = 1,2 mol\)

Gọi n_{\(Fe_2O_3\)} và n_\(CuO\) là x, y

\(\begin{cases} 160x + 80y=40\\ 6x + 2y= 1,2 \end{cases} \)

\(\begin{cases} x=0,1\\ y=0,3 \end{cases} \)

\(\Rightarrow m_{Fe_2O_3}= 0,1 . 160= 16g\)

\(m_{CuO} = 0,3 . 80=24g\)

a. \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(n_{Fe}=\frac{11,2}{56}=0,2mol\)

b. Theo phương trình \(n_{HCl}=n_{Fe}.2=0,2.2=0,4mol\)

\(\rightarrow V_{ddHCl}=\frac{0,4}{2}=0,2l=200ml\)

c. Theo phương trình \(n_{FeCl_2}=n_{Fe}=0,2mol\)

\(\rightarrow C_{M_{ddFeCl_2}}=\frac{0,2}{0,2}=1M\)

Bài 1:

Gọi KL cần tìm là A.
PT: \(A+2HCl\rightarrow ACl_2+H_2\)

Ta có: \(n_{HCl}=0,1.6=0,6\left(mol\right)\)

Theo PT: \(n_A=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{7,2}{0,3}=24\left(g/mol\right)\)

Vậy: KL cần tìm là Mg.

Bài 2:

PT: \(2R+6HCl\rightarrow2RCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{9,408}{22,4}=0,42\left(mol\right)\)

Theo PT: \(n_R=\dfrac{2}{3}n_{H_2}=0,28\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{7,56}{0,28}=27\left(g/mol\right)\)

Vậy: R là Al.

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(50ml=0,05l\)

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,05}=6\left(M\right)\)