How many grams of CaCO₃ are required to produce 2.24 L of CO₂ at STP with excess HCl? (Molar mass: CaCO₃ = 100 g/mol)
Reaction: CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O. Moles of CO₂ = 2.24 / 22.4 = 0.1 mol. 1 mol CO₂ from 1 mol CaCO₃; mass = 0.1 × 100 = 10 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Stoichiometry and Stoichiometric Calculations and Limiting Reagent
