$n_{CaCO_3(1)}=\dfrac{62}{100}=0,62(mol)$
$n_{CaCO_3(2)}=n_{Ca(HCO_3)_2}=\dfrac{50}{100}= 0,5(mol)$
Bảo toàn $C$:
$n_{CO_2}=n_{CaCO_3(1)}+2n_{Ca(HCO_3)_2}=1,62(mol)$
$(C_6H_{10}O_5)_n \to nC_6H_{12}O_6\to 2nCO_2$
$\to n_{\rm tinh bột}=\dfrac{1,62}{2n}=\dfrac{0,81}{n}(mol)$
Mà $H=80\%$, $\%m_{\rm tinh bột}=100-19=81\%$
$\to m=\dfrac{0,81}{n}.162n:80\%:81\%=202,5g$