\(a)2Na+2HCl\xrightarrow[]{}2NaCl+H_2\\ 2K+2HCl\xrightarrow[]{}2KCl+H_2 \\ b)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Na}=a,n_K=b\\ \Rightarrow\left\{{}\begin{matrix}23a+39b=6,2\\\dfrac{1}{2}a+\dfrac{1}{2}b=0,1\end{matrix}\right.\\ \Rightarrow a=b=0,1mol\\ \%_{Na}=\dfrac{0,1.23}{6,2}\cdot100=37,1\%\\ \%_K=100-37,1=62,9\%\)

1. BT klg=>mO2=2,784-2,016=0,768g

nO2=0,768/32=0,024 mol

GS KL M hóa trị n

4M                  + nO2 => 2M2On

0,096/n mol<=0,024 mol

=>0,096M=2,016n=>M=21n=>chọn n=8/3=>M=56 Fe

TN1: BT klg=>mO2=17,4-10,52=6,88g

=>nO2=0,215 mol

Quá trình nhận e

$O2$          +4e => 2 $O-2$ 

0,215 mol=>0,86 mol

n e nhận ko đổi=0,86 mol

TN2 2$H+$   +2e =>H2

0,86 mol<=0,86 mol 

Lượng KL vẫn vậy nên n e nhận ko đổi=0,86 mol=nH+=nHCl

=>VddHCl=0,86/1,25=0,688lit

Pls ai giúp mình đi cần gấp lắm

a)

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

=> 56a + 24b = 18,4 (1)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a-->2a------>a------>a

             Mg + 2HCl --> MgCl₂ + H₂

             b--->2b------->b------>b

=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (2)

(1)(2) => a = 0,2 (mol); b = 0,3 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{18,4}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,3.24}{18,4}.100\%=39,13\%\end{matrix}\right.\)

b) \(n_{HCl\left(pư\right)}=2a+2b=1\left(mol\right)\)

=> \(n_{HCl\left(tt\right)}=\dfrac{1.125}{100}=1,25\left(mol\right)\)

=> m_HCl(tt) = 1,25.36,5 = 45,625 (g)

=> \(a=\dfrac{45,625.100}{18,25}=250\left(g\right)\)

c) 

m_{dd sau pư} = 18,4 + 250 - 0,5.2 = 267,4 (g)

\(C\%_{FeCl_2}=\dfrac{0,2.127}{267,4}.100\%=9,5\%\) 

\(C\%_{MgCl_2}=\dfrac{0,3.95}{267,4}.100\%=10,66\%\)

a) Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right);n_{H_2}=\dfrac{7,28}{22,4}=0,325\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

            a-------------------------->1,5a

           Fe + 2HCl ---> FeCl₂ + H₂ 

           b---------------------------->b

b) Giải hệ PT: \(\left\{{}\begin{matrix}27a+56b=9,65\\1,5a+b=0,325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\left(TM\right)\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,15.27}{9,65}.100\%=42\%\\\%m_{Fe}=100\%-42\%=58\%\end{matrix}\right.\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\Rightarrow m_{Fe} = 0,2.56 = 11,2(gam)$
$\Rightarrow m_{Cu} = 15,6 - 11,2  = 4,4(gam)$