Đáp án:
a.23(g)
b.50(g)
Giải thích:
$n_{CO2}$=$\frac{11.2}{22.4}$=0.5(mol)
$C_{6}$$H_{12}$$O_{6}$→ 2$C_{2}$$H_{5}$$OH$ + 2$C$$O_{2}$
từ pt suy ra:
a.
$n_{CO2}$=$n_{C2H5OH}$=0.5
$m_{C2H5OH}$=n*M=0.5*46=23(g)
b.
$n_{C6H12O6}$=$\frac{1}{2}$*$n_{CO2}$=$\frac{1}{2}$*0.5=0.25
$m_{C6H12O6pư}$=n*M=0.25*180=45(g)
mà H=$\frac{mC6H12O6pư}{mC6H12O6tg}$*100=90%
⇔$\frac{45}{mC6H12O6tg}$*100=90%
⇔$m_{C6H12O6tg}$=50(g)