How many grams of `80%` pure marble stone on calination can give 14 grams of quick lime ?

Explanation

To solve the problem of how many grams of 80% pure marble stone are needed to produce 14 grams of quicklime (CaO), we can follow these steps:

### Step 1: Write the chemical reaction
The reaction for the calcination of marble stone (calcium carbonate, CaCO3) is:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

### Step 2: Determine the molar masses
- Molar mass of CaCO3 (calcium carbonate) = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol
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