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How many of 0.1N HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of two?

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How many of 0.1N HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of two?

(a) 157.7 mL

(b) 15.77 mL

(c) 147.7 mL

(d) 14.77 mL

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Correct answer: (a) 157.7 mL

Explanation:

Na2CO3    NaHCO3

x              (1 - x)

\(\frac{x}{106}\) = \(\frac{1-x}{84}\) given (moles are equal )

x = 0.557

\(\frac{0.557}{53}\) + \(\frac{0.443}{84}\) = \(\frac{V \times 0.1}{1000}\)

V = 157.7 mL

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