2.00 g of a mixture of NaHCO3 and KClO3 requires 100 mL of 0.1 N HCl for complete neutralization. The weight of KClO3 present in the mixture is:

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2.00 g of a mixture of ${NaHCO}_{3}$ and ${KClO}_{3}$ requires 100 mL of 0.1 N HCl for complete neutralization. The weight of ${KClO}_{3}$ present in the mixture is:

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0.84 g

0.16 g

1.16 g

1.84 g

answer is C.

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Detailed Solution

The weight of ${KClO}_{3}$ present in the mixture can be calculated as,

$\frac{w_{{NaHCO}_{3}}}{\frac{84}{1}} \times 1000 = 100 \times 0.1$

$\Rightarrow w_{{NaHCO}_{3}} = 0 .84 g$

Thus, option C is correct.

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