Răspuns:
M,C2H5OH=46g/mol
Vm=22,4L/mol
MASA de etanol pur, cu p=90%
p=m,px100/m,i---> m,p=100x90/100g=90g
1mol...............................................0.5mol
C2H5OH+Na---> C2H5OH+ 1/2H2
46g.................................................22,4/2l
90g........................................................x
x=V=21,9 L
Explicație: