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12.5g of commercial sodium bicarbonate gives 3 g of carbon dioxide. What is the percentage in purity in the sample? $$\ce{2NaHCO_3 -> Na_2CO_3 + CO_2 + H_2O}$$ I have solved this question, but I just need my solution verified, as it’s not given. I have solved it as follows

Molar mass of sodium bicarbonate=84

Molar mass of carbon dioxide=44

expected yield of carbon dioxide=3.27g, obtained using stoichiometric calculations.

But obtained product=3g

%yield =$\frac{obtained product}{expected product}.100$

Therefore %yield =91.74%

The answer given is 8(I don’t know how that’s possible), so I would appreciate if someone would let me know if my solution was right. Thanks!

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  • $\begingroup$ I haven't checked the numbers, but the calculations are set up correctly. Are you sure that you are not asked about the percentage of impurity? If that is the case, 100-91.74 is quite close to 8 :D $\endgroup$
    – user32223
    Aug 16, 2019 at 16:55
  • $\begingroup$ Yup, it’s asked as percentage purity $\endgroup$
    – Aditya
    Aug 17, 2019 at 10:11

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