\(4A+3O_2\rightarrow\left(t^o\right)2A_2O_3\\ m_{O_2}=32-22,4=9,6\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\ \Rightarrow n_A=\dfrac{4}{3}.0,3=0,4\left(mol\right)\\ \Rightarrow M_A=\dfrac{22,4}{0,4}=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Sắt\left(Fe=56\right)\)
-PTHH:\(4A+3O_2\rightarrow^{t^0}2A_2O_3\).
-Theo định luật bảo toàn khối lượng, ta có:
\(m_A+m_{O_2}=m_{A_2O_3}\)
\(\Rightarrow m_{O_2}=m_{A_2O_3}-m_A=32-22,4=9,6\left(g\right)\)
\(\Rightarrow n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\).
-Theo PTHH ở trên, ta có:
\(n_{A_2O_3}=\dfrac{0,3}{3}.2=0,2\left(mol\right)\)
\(\Rightarrow M_{A_2O_3}=\dfrac{m}{n}=\dfrac{32}{0,2}=160\) (g/mol).
\(\Rightarrow2.M_A+3.16=160\)
\(\Rightarrow M_A=\dfrac{160-3.16}{2}=56\) (g/mol).
\(\Rightarrow A\) là Fe (Iron).
