Limestone $(C a C O_{3})$ decompose into quicklime $(C a O)$ on stone heating. How much quantity of limestone will be required to prepare $100 k g$ of quicklime?
$(C a = 40, O = 16, C = 12)$

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Solution

$C a C O_{3} Δ - \to C a O + C O_{2} (g) ⟹ 1 m o l e o f q u i c k l i m e i s f o r m e d f r o m 1 m o l e o f C a C O_{3} . H e n c e, 56 g o f C a O = 100 g o f C a C O_{3} 1 g o f C a O = \frac{100}{56} g o f C a C O_{3} 100 g o f C a O = \frac{100 \times 100}{56} g o f C a C O_{3} = 178.57 g o f C a C O_{3} H e n c e, 178.57 k g o f l i m e s t o n e i s r e q u i r e d t o p r e p a r e 100 k g o f q u i c k l i m e .$

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