A tablet contains $2.50 g$ of $NaHCO X_{3}$ , which reacts completely with excess stomach acid: $NaHCO X_{3} + HCl NaCl + H X_{2} O + CO X_{2}$ . Taking atomic masses H = 1, C = 12, O = 16, Na = 23 and the molar volume at STP as $22.4 L mo l^{- 1}$ , calculate the volume of $CO X_{2}$ released at STP in mL. Give the answer to the nearest integer.

Numerical answer type, answer in mL, tolerance 0.5.

Answer: 667 mL

Moles of NaHCO3 = 2.50/84 = 0.02976 mol, and 1 mol gives 1 mol CO2. Volume = 0.02976 x 22.4 L = 0.6667 L = 666.67 mL, which is 667 mL to the nearest integer.

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Answer: 667 mL

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