Question

calcium carbonate is 90% pure.
volume of carbon dioxide is collected at STP when 10 gram of calcium carbonate is decomposed is​

Answer 1

CaCO3 is 90% pure

Mass 10*90/100

= 9 gm

n= 9/100

Volume of CO2 =9*22.4/100

=2.016 l

Answer 2

Answer:

The volume of CO₂ produce will be 2.016L

Explanation:

• CaCO₃  ⇄  CO₂ + CaO

• 1 mol of calcium carbonate produce one mol of carbon dioxide on decay.
• Molecular weight of calcium carbonate is 40 + 12 + 48 = 100.
• 10 gm of CaCO₃ forms 10/100 = 0.1 mol in 100% pure form will produce 0.1 gm CO₂.
• 90% of 0.1gm would be 0.09mol that would produce 0.09 mol of CO₂.
• At STP volume is 22.4 L for 1 mol. So, the volume at 0.09 mol will be 0.09(22.4) = 2.016 L

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