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

Phương trình hóa học

Mg + 2HCl ---> MgCl₂ + H₂ (**)

MgO + 2HCl ---> MgCl₂ + H₂O (***)

b) Từ (*) và (**) ta có \(n_{Mg}=0,15\Leftrightarrow m_{Mg}=0,15.24=3,6\left(g\right)\)

\(\Rightarrow m_{MgO}=10-3,6=6,4\left(g\right)\)

 \(\%Mg=\dfrac{3,6}{10}.100\%=36\%\)

\(\%MgO=\dfrac{6,4}{10}.100\%=64\%\)

c) Xét phản ứng (**) ta có \(m_{MgO}=6,4\left(g\right)\Leftrightarrow n_{MgO}=n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,16\left(mol\right)\) (1) 

\(\Leftrightarrow n_{HCl}=0,32\left(mol\right)\)

Tương tự có số mol HCl trong phản ứng (*) là 0,3 mol

\(C_M=\dfrac{0,32+0,3}{0,2}=3,1\left(M\right)\)

d) Từ (1) ; (*) ; (**) ta có : \(n_{MgCl_2}=0,15+0,16=0,31\left(mol\right)\)

\(m_{MgCl_2}=0,31.95=29,45\left(g\right)\)

e) \(C_M=\dfrac{0,31}{0,2}=1,55\left(M\right)\)

Có lẽ đề hỏi %m mỗi chất bạn nhỉ?

Ta có: 46n_C2H5OH + 60n_CH3COOH = 21,2 (1)

PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

\(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{CH_3COOH}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,2.46}{21,2}.100\%\approx43,4\%\\\%m_{CH_3COOH}\approx56,6\%\end{matrix}\right.\)

\(Đặt:n_{C_6H_5OH}=a\left(mol\right);n_{CH_3OH}=b\left(mol\right)\left(a,b>0\right)\\ C_6H_5OH+Na\rightarrow C_6H_5ONa+\dfrac{1}{2}H_2\\ CH_3OH+Na\rightarrow CH_3ONa+\dfrac{1}{2}H_2\\ \Rightarrow\left\{{}\begin{matrix}94a+32b=26,7\\22,4.0,5a+22,4.0,5b=3,92\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,25\\b=0,1\end{matrix}\right.\\ \Rightarrow m_{CH_3OH}=0,1.32=3,2\left(g\right)\\ \Rightarrow m_{C_6H_5OH}=94.0,25=23,5\left(g\right)\)

PTHH: 

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1) 

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

\(n_{HCl\left(1\right)+\left(2\right)}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

\(n_{H_2}=\dfrac{7,437}{24,79}0,3\left(mol\right)\)

Theo PTHH (1): \(n_{Al}=\dfrac{2\cdot0,3}{3}=0,2\left(mol\right)\); \(n_{HCl\left(1\right)}=\dfrac{6\cdot0,3}{3}=0,6\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2\cdot27=5,4\left(g\right)\)  

\(\Rightarrow n_{HCl\left(2\right)}=1-0,6=0,4\left(mol\right)\)

Theo PTHH (2): \(n_{CuO}=\dfrac{1\cdot0,4}{2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)

\(\Rightarrow m_A=m_{CuO}+m_{Al}=16+5,4=21,4\left(g\right)\)

% khối lượng của mỗi chất trong hỗn hợp A là: 

\(\%m_{Al}=\dfrac{5,4\cdot100}{21,4}\approx25,2\%\)

\(\Rightarrow\%m_{CuO}=100\%-25,2\%=74,8\%\)

a, Ta có: 65n_Zn + 27n_Al = 11,9 (1)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

⇒ m_Zn = 0,1.65 = 6,5 (g)

m_Al = 0,2.27 = 5,4 (g)

b, Theo PT: n_ZnCl2 = n_Zn = 0,1 (mol)

n_AlCl3 = n_Al = 0,2 (mol)

⇒ m _muối = 0,1.136 + 0,2.133,5 = 40,3 (g)

c, Theo PT: n_HCl = 2n_H2 = 0,8 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{0,8.36,5}{10\%}=292\left(g\right)\)

a. PTHH 1 : Ba + H₂O -> Ba(OH)₂ + H₂

                   0,025                             0,025

PTHH 2 : BaO + H₂O -> Ba(OH)₂ 

\(n_{H_2}=\dfrac{0.56}{22,4}=0,025\left(mol\right)\)

\(m_{Ba}=0,025.137=3,425\left(g\right)\)

\(m_{BaO}=6,485-3,425=3,06\left(g\right)\)

\(n_{BaO}=\dfrac{3.06}{153}=0,02\left(mol\right)\)

b. \(\%m_{BaO}=\dfrac{3,06}{6,485}.100=47,2\%\)

\(\%m_{Ba}=100\%-47,2\%=52,8\%\)

a

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

0,15<-0,15<--0,15<----0,15

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

0,16-->0,32---->0,16

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

\(m_{Mg}=0,15.24=3,6\left(g\right)\\ m_{MgO}=10-3,6=6,4\left(g\right)\)

b

\(\%m_{Mg}=\dfrac{3,6.100\%}{10}=36\%\\ \%m_{MgO}=\dfrac{6,4.100\%}{10}=64\%\)

c

\(n_{MgO}=\dfrac{6,4}{40}=0,16\left(mol\right)\)

\(CM_{HCl}=\dfrac{0,15+0,32}{0,2}=2,35M\)

d

\(m_{MgCl_2}=\left(0,15+0,16\right).95=29,45\left(g\right)\)

e

\(CM_{MgCl_2}=\dfrac{0,15+0,16}{0,2}=1,55M\)

a) Theo đề bài: \(m_{Cu}=0,3\left(g\right)\)

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

               a______2a______a_____a         (mol)

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

               b_____2b______b____b              (mol)

Ta lập được hệ phươn trình: \(\left\{{}\begin{matrix}24a+56b=18,7-0,3=18,4\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,3\cdot24=7,2\left(g\right)\\m_{Fe}=0,2\cdot56=11,2\left(g\right)\end{matrix}\right.\)

b) PTHH: \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2+2H_2O\)

Ta có: \(n_{Cu}=\dfrac{0,3}{64}=\dfrac{3}{640}\left(mol\right)=n_{SO_2}\) \(\Rightarrow V_{SO_2}=\dfrac{3}{640}\cdot22,4=0,105\left(l\right)\)

a)

Gọi $n_{Mg} = a(mol) ; n_{Fe} = b(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{H_2} = a + b = \dfrac{11,2}{22,4} = 0,5(mol)$
$m_X = 24a + 56b = 0,3 = 18,7(gam)$
Suy ra a = 0,3 ; b = 0,2

$m_{Mg} = 0,3.24 = 7,2(gam)$

$m_{Fe} = 0,2.56 = 11,2(gam)$

b)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + H_2O$
$n_{SO_2} = n_{Cu} = \dfrac{0,3}{64}$

$V = \dfrac{0,3}{64}.22,4 = 0,105(lít)$