\(4.\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

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

\(0.15.....0.3....................0.15\)

\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.3}{0.5}=0.6\left(M\right)\)

\(5.\)

\(Đặt:n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_{hh}=56a+27b=8.3\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

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

\(\Rightarrow a+1.5b=0.25\left(2\right)\)

\(\left(1\right),\left(2\right):a=b=0.1\)

\(\%Fe=\dfrac{5.6}{8.3}\cdot100\%=67.47\%\)

\(\%Al=32.53\%\)

nFe = 0,1 mol. m O2 = 7,36 - 5,6 = 1,76 gam => n O2 = 0,055 mol
Đặt nNO = x, nNO2 = y
Ta có hệ: bảo toàn electron: 3.0,1 - 0,055.4 = 3x + y
30x + 44y / ( x+y) = 19.2
=> x= 0,02, y =0,02
=> V = 0,04.22,4 = 8,96 lít

A.2,24 lít

\(CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O\)

\(n_{CaCO_3}=\dfrac{10}{100}=0,1 mol\)

Theo PTHH:

\(n_{CO_2}=n_{CaCO_3}= 0,1 mol \)

\(\Rightarrow V_{CO_2}= 0,1 . 22,4= 2,24 l\)

Fe2O3+3H2---->2Fe+ 3H2O

x-----------3x--------2x

CuO+H2--->Cu+H2O

y------y---------y

Gọi n Fe2O3 =x suy ra m Fe =112x

n CuO =y-----> mCu=64y

n\(_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

Theo bài ra ta có pt

\(\left\{{}\begin{matrix}3x+y=0,5\\112x+64y=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

m =m \(_{Fe2O3}+m_{CuO}=0,1.160+0,2.80=32\left(g\right)\)

Ta có : nH2=0,5(mol)

Gọi a , y lần lượt là số mol của Fe2O3 , CuO

PTHH: Fe2O3+3H2--->2Fe+3H2O

CuO+H2--->Cu+H2O

Tự đánh dấu tỉ lệ nhé

\(\Rightarrow\left\{{}\begin{matrix}3x+y=0,5\\112x+64y=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=>mdd=32g

a)

\(n_{O_2} = \dfrac{5,6}{22,4} = 0,25(mol)\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)

Theo PTHH :

\(n_{Fe} = \dfrac{3}{2}n_{O_2} = 0,375(mol)\\\Rightarrow a = 0,375.56 = 21(gam)\)

b)

Theo PTHH :

\(n_{Fe_3O_4} = 0,5n_{O_2} = 0,25.0,5 = 0,125(mol)\\\Rightarrow m_{Fe_3O_4} = 0,125.232 = 29(gam)\)