How many grams of Na₂CO₃ are required to produce 8.4 g of NaCl with excess HCl? (Molar masses: Na₂CO₃ = 106 g/mol, NaCl
Reaction: Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O. Moles of NaCl = 8.4/58.5 ≈ 0.1436 mol. 2 mol NaCl from 1 mol Na₂CO₃; 0.1436 mol from 0.0718 mol. Mass = 0.0718 × 106 ≈ 7.61 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Reactions in Solutions and Mass Percentage and Numericals