\(n_{HCl}=0,4.0,2=0,08\left(mol\right)\\PTHH:\)

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

x------->2x----->x

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

y------->2y------>y

Có hệ PT: \(\left\{{}\begin{matrix}80x+40y=2,8\\2x+2y=0,08\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,03\\y=0,01\end{matrix}\right.\)

a)

\(m_{CuO}=80.0,03=2,4\left(g\right)\\ m_{MgO}=40.0,01=0,4\left(g\right)\)

b)

\(\%_{m_{CuO}}=\dfrac{2,4.100\%}{2,8}=85,71\%\\ \%_{m_{MgO}}=\dfrac{0,4.100\%}{2,8}=14,29\%\)

c)

\(m_{CuCl_2}=135x=135.0,03=4,05\left(g\right)\)

a)\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,2                                        0,1

\(NaO+H_2O\rightarrow2NaOH\)

b)\(m_{Na}=0,2\cdot23=4,6g\)

   \(m_{NaO}=m_{hh}-m_{Na}=40,5-4,6=35,9g\)

c)\(n_{NaO}=\dfrac{35,9}{39}=0,92mol\Rightarrow n_{NaOH}=2n_{NaO}=1,84mol\)

   \(\Rightarrow m_{NaOH}=1,84\cdot40=73,6g\)

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

a.

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

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

b.

\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right),m_{CuO}=12-2,4=9,6\left(g\right)\)

c. 

\(m_{muối}=m_{CuCl_2}+m_{MgCl_2}=0,12.135+95.0,1=25,7\left(g\right)\)

Em bổ sung khối lượng hỗn hợp ban đầu nhé !

\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

            1          2             1          1

          0,2       0,4                       0,2

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

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(m_{Cu}=16-11,2=4,8\left(g\right)\)

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

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

 Chúc bạn học tốt

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe_2O_3}=a\left(mol\right);n_{CuO}=b\left(mol\right)\left(a,b>0\right)\\ m_{hhoxit}=k\left(g\right)\\ \Rightarrow\left(1\right)160a+80b=k\\ \left(2\right)112a+64b=0,72k\\ \Rightarrow6,4a=12,8b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{12,8}{6,4}=\dfrac{2}{1}\\ \Rightarrow\%m_{Fe_2O_3}=\dfrac{160.2}{160.2+80.1}.100=80\%\\ \Rightarrow\%m_{CuO}=20\%\)

\(2Mg+O_2-^{t^o}\rightarrow2MgO\\ 2Cu+O_2-^{t^o}\rightarrow2CuO\\ Đặt:\left\{{}\begin{matrix}m_{Mg}=x\left(g\right)\\m_{Cu}=y\left(g\right)\end{matrix}\right.\\\Rightarrow\left\{{}\begin{matrix}n_{Mg}=\dfrac{x}{24}\left(mol\right)\\n_{Cu}=\dfrac{x}{64}\left(mol\right)\end{matrix}\right.\\ TheoPT:\Rightarrow\left\{{}\begin{matrix}n_{MgO}=\dfrac{x}{24}\left(mol\right)\\n_{CuO}=\dfrac{x}{64}\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{24}.40=25\%.\left(\dfrac{x}{24}.40+\dfrac{y}{64}.80\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=12\end{matrix}\right.\)

Đặt n_Al = a (mol)

=> n_Mg = 2a (mol)

=> 27a + 24 . 2a = 15

=> a = n_Al = 0,2 (mol)

n_Mg = 0,2 . 2 = 0,4 (mol)

m_Al = 0,2 . 27 = 5,4 (g)

m_Mg = 0,4 . 24 = 9,6 (g)

%m_Al = \(\dfrac{5,4}{15}=36\%\)

%m_Mg = 100% - 36% = 64%

PTHH:

2Al + 6HCl -> 2AlCl₃ + 3H₂

0,2 ---> 0,6 ---------------> 0,3

Mg + 2HCl -> MgCl₂ + H₂

0,4 ---> 0,8 --------------> 0,4

=> V_H2 = (0,3 + 0,4) . 22,4 =15,68 (l)

=> m_HCl = (0,6 + 0,8) . 36,5 = 51,1 (g)

Bài 3 :

\(a) n_{CuO} = a(mol) ; n_{Fe_2O_3} = b(mol)\\ \Rightarrow 80a + 160b = 36(1)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + H_2O\\ n_{Cu} = n_{CuO} = a(mol)\\ n_{Fe} = 2n_{Fe_2O_3} = 2b(mol)\\ \Rightarrow 64a = 4.2b.56(2)\\ (1)(2) \Rightarrow a = 0,35 ; b = 0,05\\ m_{CuO} = 0,35.80 = 28(gam)\\ m_{Fe_2O_3} = 0,05.160 = 8(gam)\\ b) n_{H_2} = a + 3b = 0,5(mol) \Rightarrow V_{H_2} = 0,5.22,4 = 11,2(lít)\)

\(c) Fe + 2HCl \to FeCl_2 + H_2\\ n_{HCl} = 2n_{Fe} = 0,1.2 = 0,2(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{10,95\%} = 66,67(gam)\)

Bài 4 :

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = n_{Zn} = \dfrac{1,95}{65} = 0,03(mol)\\ V_{H_2} = 0,03.22,4= 0,672(lít)\\ b) n_{HCl} =2 n_{H_2} = 0,06(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,06.36,5}{120}.100\% = 1,825\%\\ m_{dd\ sau\ pư} = 1,95 + 120 - 0,03.2 = 121,89(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,03.136}{121,89}.100\% = 3,35\%\)