a, \(C_6H_{12}O_6\underrightarrow{^{t^o,xt}}2C_2H_5OH+2CO_2\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
b, \(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)
Theo PT: \(n_{C_2H_5OH}=n_{CO_2}=n_{CaCO_3}=0,5\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,5.46=23\left(g\right)\)
\(n_{C_6H_{12}O_6\left(LT\right)}=\dfrac{1}{2}n_{CO_2}=0,25\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{C_6H_{12}O_6\left(TT\right)}=\dfrac{0,25}{80\%}=0,3125\left(mol\right)\)
\(\Rightarrow m_{C_6H_{12}O_6}=0,3125.180=56,25\left(g\right)\)