\(n_{C_2H_5OH}=\dfrac{14}{46}=\dfrac{7}{23}\left(mol\right)\)
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
\(\dfrac{7}{23}...................\dfrac{7}{23}......\dfrac{7}{46}\)
\(m_{C_2H_5ONa}=\dfrac{7}{23}\cdot68=20.7\left(g\right)\)
\(V_{H_2}=\dfrac{7}{46}\cdot22.4=3.4\left(l\right)\)
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\(a) 2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2\\ n_{C_2H_5ONa} = n_{C_2H_5OH} = \dfrac{14}{46} = \dfrac{7}{23}(mol)\\ m_{C_2H_5ONa} = \dfrac{7}{23}.68 = 20,7(gam)\\ n_{H_2} = \dfrac{1}{2}n_{C_2H_5OH} = \dfrac{7}{46}(mol)\\ m_{H_2} = \dfrac{7}{46}.2 = \dfrac{7}{23}(gam)\\ b) V_{H_2} = \dfrac{7}{46}.22,4 = 3,41(lít)\)
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