A 0.500 g vitamin C tablet (ascorbic acid, molar mass 176 g mol^-1) is titrated with 0.0500 M iodine, C6H8O6 + I2 -> C6H6O6 + 2HI, requiring 22.8 mL. The mass percent of ascorbic acid in the tablet is

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A $0.500 g$ vitamin C tablet (ascorbic acid, molar mass $176 g mo l^{- 1}$ ) is titrated with $0.0500 M$ iodine, $C X_{6} H X_{8} O X_{6} + I X_{2} C X_{6} H X_{6} O X_{6} + 2 HI$ , requiring $22.8 mL$ . The mass percent of ascorbic acid in the tablet is

The reaction is 1:1, so mol ascorbic acid = 0.0228 x 0.0500 = 1.14e-3. Mass = 1.14e-3 x 176 = 0.2006 g; percent = 0.2006/0.500 x 100 = 40.13%.