\(V_{C_2H_5OH}=\dfrac{36.92}{100}=33,12\left(ml\right)\)
=> \(m_{C_2H_5OH}=D_{C_2H_5OH}.V_{C_2H_5OH}=0,8.33,12=26,496\left(g\right)\)
=> \(n_{C_2H_5OH}=\dfrac{26,496}{46}=0,576\left(mol\right)\)
\(V_{H_2O}=36-33,12=2,88\left(ml\right)\)
=> \(m_{H_2O}=D_{H_2O}.V_{H_2O}=1.2,88=2,88\left(g\right)\)
=> \(n_{H_2O}=\dfrac{2,88}{18}=0,16\left(mol\right)\)
PTHH: 2C2H5OH + 2Na --> 2C2H5ONa + H2
0,576------------------>0,576
2H2O + 2Na --> 2NaOH + H2
0,16---------------->0,16
=> \(m_{rắn}=m_{C_2H_5ONa}+m_{NaOH}=0,576.68+0,16.40=45,568\left(g\right)\)
\(V_{C_2H_5OH}=\dfrac{36.92}{100}=33,12\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=\dfrac{33,12}{0,8}=41,4\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{41,4}{46}=0,9\left(mol\right)\)
\(PTHH:2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\uparrow\\ Mol:0,9\rightarrow0,9\rightarrow0,9\\ \rightarrow m_{C_2H_5ONa}=0,9.68=61,2\left(g\right)\)
