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in JEE by (189 points)

2.0 g of a mixture of carbonate, bicarbonate and chloride of sodium on heating produced 56 mL of CO2 at NTP. 1.6 g of the same mixture required 25 mL of N HCl solution for neutral ization. Calculate percentage of each component present in mixture.

1 Answer

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Best answer

On heating NaHCO3 decomposes to gives CO2

2NaHCO3→Na2CO3+H2O+CO2↑

Meq of NaHCO3=Meq of CO2

(wNaHCO3×1000)/84 = (56/22400)×2×1000

wNaHCO3=0.42g in 2.0g mixture

⇒0.42/2 × 100

⇒21%

Now suppose in 1.6g mixture,'x'g of Nacl,0.336g(21% of 1.6)NaHCO3 and rest Na2CO3 then

Wt of Na2CO3=1.6−0.336−x

⇒(1.264−x)g

Since both Na2CO3 and NaHCO3 react with HCl and therefore

Meq of Na2CO3+Meq of NaHCO3=Meq of HCl

Calculating for x

x=0.151

Weight of NaCl=0.151g and % of NaCl=9.43%

Weight of NaHCO3=0.336g and % of NaHCO3=21.00%

Weight of Na2CO3=1.264−0.151

⇒1.113 and % of Na2CO3=69.57%

by (52.9k points)
Awesome answer:)

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