a) PTPU
$CaCO_3 -> CaO + CO_2$
$n_{CaCO_3} = 10^6 : 100 = 10^4$ (mol)
$m_{CaO} = 10^4.56 = 560000 (g) =0,56 (tan)$
b) $m_{CaCO_3} = 1.80% = 0,8 (tan) = 8.10^5 (g)$
$n_{CaCO_3} = 8.10^5 :100 = 8.10^3 (mol)$.
$n_{CaO} = 8.10^3 .85% = 6,8.10^3 (mol)$.
$m_{CaO} = 6,8.10^3 . 56 = 380,8.10^3 (g) =0,3808 (tan)$.