Dặt: nCH3OH= x mol
nC2H5OH= y mol
mhh= 32x + 46y= 7.8 g (1)
CH3OH + Na --> CH3ONa + 1/2H2
x_________________________0.5x
C2H5OH + Na --> C2H5ONa + 1/2H2
y__________________________0.5y
nH2= 0.5x + 0.5y=0.1 mol
<=> x + y= 0.2 (2)
Giải (1) và (2)
x=y=0.1
mCH3OH= 32*0.1=3.2g
mC2H5OH= 0.1*46=4.6g
%CH3OH= 3.2/7.8*100%= 41.02%
%C2H5OH= 100-41.02=58.98%