\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
          0,3      0,6                        0,3 
\(m_{HCl}=0,6.36,5=21,9\left(g\right)\\ m_{\text{dd}HCl}=\dfrac{21,9.100}{7,3}=300\left(g\right)\)(g) 

=))

\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\\ m_{\text{dd}NaCl}=\dfrac{5,85.100}{10}=58,5\left(g\right)\)

gọi n_Fe : x , n_O : y 

ta có :
\(\dfrac{56x}{16y}=\dfrac{70}{30}\\ \Rightarrow\dfrac{x}{y}=\dfrac{70.16}{30.56}=\dfrac{2}{3}\) 
tỉ lệ = 2:3 
=> C

\(4K+O_2\underrightarrow{t^o}2K_2O\\ K_2O+H_2O\rightarrow2KOH\\ 2KOH+CO_2\rightarrow K_2CO_3+H_2O\\ K_2CO_3\underrightarrow{t^o}K_2O+CO_2\\ CO_2+CaO\rightarrow CaCO_3\\ b,2KClO_3\xrightarrow[xtMnO_2]{t^o}2KCl+3O_2\\ O_2+2Cu\underrightarrow{t^o}2CuO\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\\ Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)

\(m_{HCl}=\dfrac{7,3.200}{100}=14,6\left(g\right)\\ n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ pthh:CuO+2HCl\rightarrow CuCl_2+H_2O\) 
           0,2         0,4         0,2           0,2 
\(m_{CuO}=0,2.80=16\left(g\right)\\ m_{dd}=16+200-\left(0,2.18\right)=212,4\left(g\right)\\ C\%_{CuCl_2}=\dfrac{0,2.135}{212,4}.100\%=12,711\%\)

\(Cr+2F_2\rightarrow CrF_4\)
A