Question

how much marble of 96.5% purity would be required to prepare 10 litre of carbon dioxide at STP when the marble is acted upon by dilute hydrochloric acid​

Answer 1

Answer:

46.26 g

Explanation:

Marble is chemically CaCO₃

The reaction of CaCO₃ with HCl is given by the following equation:

CaCO₃ + 2 HCl ⇒ CaCl₂ + H₂O + CO₂

From the equation, we can conclude that

1 mole of CO₂ requires 1 mole of CaCO₃

⇒ 22.4 liter of CO₂ requires 100 g of CaCO₃  (1 mole CaCO₃ = 100 g of CaCO₃)

⇒ 10 liter of CO₂ requires (100/22.4) x 10 g of CaCO₃ = 44.64 g of CaCO₃

Since the sample is 96.5% pure, we need

44.64/0.965 = 46.26 g of marble to account for impurity