Question

calculate the weight caco3 is required to produce 5.6lit of co2 gas at stp ​

Answer 1

Answer :

We have to find amount of CaCO₃ required to produce 5.6l of CO₂ at STP$.$

⧪ Chemical reaction :

$\dag\:\boxed{\bf{CaCO_3+\Delta\longrightarrow CaO+CO_2}}$

It is clear from the reaction that, 1 mole of CaCO₃ produces 1 mole (22.4l) of CO₂ at STP.

Molar mass of CaCO₃ = 100g/mol

22.4l of CO₂ requires = 100g of CaCO₃

5.6l of CO₂ requires = (?)

• W = 5.6×100/22.4
• W = 100/4
• W = 25g

Hence, 25g of CaCO₃ is required to produce 5.6l of CO₂ at STP.

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