PT : \(M_xO_y+yCO\rightarrow xM+yCO_2\) ( 1 )
Ca(OH)\(_2\) + CO\(_2\) \(\rightarrow\) CaCO\(_3\) +H\(_2\)O ( 2 )
\(n_{CaCO_3}=\frac{3}{100}=0.03mol\)
Theo PT ( 2 ), ta có:
\(\Rightarrow n_O\)trong oxit kim loại = \(n_{CO_2}=n_{CaCO_3}=0.03mol\)
\(\rightarrow\) \(\Rightarrow m_O=0.03\cdot16=0.48g\)
Mà \(m_M=1,74-m_O=1.26g\)
\(n_{H_2}=\frac{0.504}{22.4}=0.0225mol\)
PT: 2M + 2n HCl \(\rightarrow\) 2MCl\(_n\) + nH\(_2\)
\(\frac{0.045}{n}\) mol \(\leftarrow\) 0.0225mol
M = \(1,26\div\frac{0.045}{n}=28n\)
\(\Rightarrow\) n = 2 và M=56
\(\Rightarrow\) M là Fe ( Sắt )
