2Câu 1
c6h12o6 (lên men)----> 2c2h5oh + 2co2
c2h5oh + o2 (giấm men)--------> ch3cooh + h2o
ch3cooh + na2co3-----> ch3coona + h2o + co2
co2 + naoh dư-----> na2co3 + h2o
Câu 2
n hcl= 0.3 *2 =0.6(mol)
cuo + 2hcl-----> cucl2 + h2o
x 2x
mgo + 2hcl-----> mgcl2 + h2o
y 2y
Gọi n cuo= x, n mgo= y
Có
80x + 40y= 16
2x + 2y = 0.6
=) x= 0.1, y= 0.2
=) m cuo= 0.1*80=8(g)
m mgo= 16-8=8(g)
3/
n c2h5oh= (9.2%*5000*0.8)/46=8(mol)
c2h5oh + o2 (giấm men)--------> ch3cooh + h2o
8 8
Do H= 80%
=) m ch3cooh= 8*80%*60=384(g)
