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A 200 g mixture of Na2CO3, NaHCO3, and Na2SO4 is heated until CO2 evolution ceases. The CO2 collected measures 3 L at 2.5 atm and 300 K. The entire original sample requires 1.5 L of 1 M HCl for complete neutralisation. Find the mass percentage of Na2CO3 in the mixture. (Na = 23, S = 32; R = 1/12 L atm mol⁻¹ K⁻¹)

1. 10%
2. 20%
3. 30%
4. 40%

Correct answer: 30%

Solution

Moles CO2 = PV/RT = (2.5*3)/(300*(1/12)) = 7.5/25 = 0.3 mol. So y = 2*0.3 = 0.6 mol NaHCO3. HCl: 2x + 0.6*1 = 1.5 => 2x = 0.9 => x = 0.45 mol Na2CO3. Mass Na2CO3 = 0.45*106 = 47.7 g. % = 47.7/200 * 100 = 23.85%. Closest option is 20% but exam key says 30%. Let me try: HCl neutralises the sample BEFORE heating, so NaHCO3 also present: 2x + y = 1.5 => 2x + 0.6 = 1.5 => x = 0.45. Same. Trying another: maybe the question says HCl neutralises only the HEATED residue (which contains Na2CO3 from original + from NaHCO3 decomposition). 2*(x+0.3) = 1.5 => x+0.3 = 0.75 => x = 0.45. Same. Mass = 47.7 g, ~ 24%. Given options, going with 20% seems closest numerically but 30% may be the exam key.

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