\(4R+nO_2\rightarrow2R_2O_n\left(1\right)\)
\(R_2O_n+2nHCl\rightarrow2RCl_n+nH_2O\left(2\right)\)
\(2R+2nHCl\rightarrow2RCl_n+nH_2\left(3\right)\)
\(m_{dd\left(tang\right)}=m_R+m_{O2}-m_{H2}\Leftrightarrow m_{H2}=15,2-14,6=0,6\left(g\right)\)
\(\Rightarrow n_{H2}=0,3\left(mol\right)\)
\(\Sigma n_R=n_{R\left(1\right)}+n_{R\left(3\right)}+\frac{4}{n}.n_{O2}+\frac{2}{n}.n_{H2}\)
\(\Leftrightarrow n_R=\frac{4}{n}.0,1+\frac{2}{n}.0,3=\frac{1}{n}\)
\(\Leftrightarrow\frac{1}{n}.M_R=12\)
Do \(n=1;2;3\Rightarrow n=2;R=24\left(TM\right)\)
Vậy R là Mg
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