Ta có: \(n_{CO_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
PT: \(C_6H_{12}O_6\underrightarrow{^{t^o,xt}}2C_2H_5OH+2CO_2\)
_______0,5___________1_______1 (mol)
a, \(m_{C_2H_5OH}=1.46=46\left(g\right)\)
b, \(m_{C_6H_{12}O_6\left(LT\right)}=0,5.180=90\left(g\right)\)
Mà: H = 80%
\(\Rightarrow m_{C_6H_{12}O_6\left(TT\right)}=\dfrac{90}{80\%}=112,5\left(g\right)\)