\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(R+2HCl\rightarrow RCl_2+H_2\)
\(0.1...............................0.1\)
\(M_R=\dfrac{5.6}{0.1}=56\left(\dfrac{g}{mol}\right)\)
\(R:Fe\left(Sắt\right)\)
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