The amount of CaCO₃ that are consumed at the reaction is 6.7 grams.
Ideal gas equation will be represented PV = nRT from which we can calculate the moles of carbon dioxide, where
On putting all these values, we get
n = (1)(1.5) / (0.082)(273) = 0.0669 moles = 0.067 mol
Given chemical equation is:
CaCO₃ + 2HCl → CO₂ + CaCl₂ + H₂O
From the stoichiometry of the reaction it is clear that,
1 moles of CO₂ = produced by 1 moles of CaCO₃
0.067 moles of CO₂ = produced by 0.067 moles of CaCO₃
Now mass of CaCO₃ will be calculated as:
n = W/M, where
W = (0.067)(100) = 6.7g
Hence required mass of CaCO₃ is 6.7g.
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