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Weight of 40% pure limestone which liberates 4.4 grams of carbon dioxide on heating is

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25 g

60 g

1 g

40 g

answer is A.

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Detailed Solution

CaCO₃(s) →CaO(s) + CO₂(g)
100g 44 g
10g 4.4 g
weight of 100% pure limestone required = 10g
weight of 40% pure limestone required = $10 \times \frac{100}{40} = 25 g;$

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