Đáp án A
${{\left( {{C}_{6}}{{H}_{10}}{{O}_{5}} \right)}_{n}}\text{ }\xrightarrow{len\,men}\text{ }2nC{{O}_{2}}\text{ }+\text{ }2n{{C}_{2}}{{H}_{5}}OH$
$C{{O}_{2}}\text{ }+\text{ }Ca{{\left( OH \right)}_{2}}\text{ }\to \text{ }CaC{{O}_{3}}\text{ }+\text{ }{{H}_{2}}O$
${{n}_{C{{O}_{2}}}}={{n}_{CaC{{O}_{3}}}}=0,75\text{ }mol$
${{m}_{tinh\,bot}}=\frac{{{m}_{tinh\,bot\left( pu \right)}}}{H%}=\frac{\frac{1}{2n}{{n}_{C{{O}_{2}}}}.162n}{H%}=\frac{0,375.162}{81%}=75\text{ }gam$.
