Question

Answer 1

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,3<-------------------0,3

=> m_Fe = 0,3.56 = 16,8 (g)

=> m_Cu = 23,2 - 16,8 = 6,4 (g)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{16,8}{23,2}.100\%=72,414\%\\\%m_{Cu}=\dfrac{6,4}{23,2}.100\%=27,586\%\end{matrix}\right.\)

Answer 2

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

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

0,3                                  0,3

\(m_{Fe}=0,3\cdot56=16,8g\)

\(\%m_{Fe}=\dfrac{16,8}{23,2}\cdot100\%=72,41\%\)

\(\%m_{Cu}=100\%-72,41\%=27,59\%\)

Answer 3

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

nFe = 0,3 (mol)

mFe = 0,3 . 16,8 (g)

%mFe = 16,8/23,2 = 72,41%

%mCu = 100% - 72,41% = 27,59%

Answer 4

Fe+HCl->Fecl2+H2

0,3-------------------0,3 mol

n H2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>m Fe=0,3.56=16,8g=>m Cu=6,4g

=>%m Fe=\(\dfrac{16,8}{23,2}\).100=72,41%

=>%m Cu=27,59%

Answer 5

⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩%mFe=16,823,2.100%=72,414%%mCu=6,423,2.100%=27,586%