Weight of 20% pure limestone which liberates 4.4 grams of carbon dioxide on heating is

CaCO₃(s) →CaO(s) + CO₂(g)
100g                              44 g
10g                                4.4 g
weight of 100% pure limestone required = 10g
weight of 20% pure limestone required   = $10\times \frac{100}{20}=50 \mathrm{g} ;$