\(V_{C_2H_5OH\left(nguyên.chất\right)}=\dfrac{0,5.30}{100}=0,15l\)
\(0,15lít=150ml\)
\(V_{H_2O}=500-150=350ml\)
\(m_{C_2H_5OH\left(nguyên.chất\right)}=150.0,8=120g\)
\(m_{H_2O}=350.1=350g\)
\(n_{C_2H_5OH}=\dfrac{120}{46}=2,6mol\)
\(n_{H_2O}=\dfrac{350}{18}=19,44mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
2,6 1,3 ( mol )
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
19,44 9,72 ( mol )
\(V_{H_2}=\left(1,3+9,72\right).22,4=246,848l\)
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