68.C
$(C_6H_{10}O_5)_n+nH_2O\xrightarrow{enzim}nC_6H_12O_6$
$C_6H_{12}O_6\xrightarrow{\text{men rượu}}2CO_2+2C_2H_5OH$
$Ca(OH)_2+CO_2\xrightarrow{}CaCO_3+H_2O$
$n_{CaCO_3}=n_{CO_2}=\dfrac{75}{100}=0,75(mol)$
$n_{CO_2}=2n_{C_6H_{12}O_6}=2n_{C_6H_{10}O_5}\text{lý thuyết}$
$⇒n_{C_6H_{10}O_5}\text{lý thuyết}=\dfrac{0,75}{2}=0,375(mol)$
$⇒n_{C_6H_{10}O_5}\text{thực tế}=\dfrac{0,375}{0,81.0,81}=\dfrac{3250}{6561}(mol)$
$m_{C_6H_{10}O_5}=\dfrac{3250}{6561}.162≈80(g)$
$⇒C$
69.
$C_{12}H_{22}O_{11}+H_2O\xrightarrow{H_2SO_4}C_6H_{12}O_6+C_6H_{12}O_6$
$n_{C_{12}H_{22}O_{11}}=\dfrac{68,4}{324}=0,2(mol)$
$n_{X}\text{thực tế}=0,2.2.0,92=0,368(mol)$
$n_X=2n_{Ag}$ ( gồm tráng bạc của Glucozo và Fructozo luôn)
$⇒m_{Ag}=0,368.2.108=79,488(g)$
$⇒A$
