Question

i need help with the bonus question

CO 498 Energy (kJ/mol) .C=O bond in CO2 CEO 799* 1072 L2 [C=o] + 0 80] -2 Co=c=0) Bonus question: 2 (1072) + 498-2(Z (799)) -> (554) The active ingredient in the antacid Tums is calcium carbonate, Caco, What volume of 0.15 M stomach acid (HCH could be neutralized by a dose of two tablets, each containing 500.0 mg of the active ingredient? --> 1000 mg MM of CaCO₃=100.0869 g/mol

Answer 1

00.133 ml stomach acid (HCl) could be neutralised by a dose of two tablets

Reaction: CaCO₃ + Que Calle + H₂O + CO2 Molecular mass of CaCO₃ = 100.08 69 g/mol Mass of Caco; given (2 tablets) 2x 500 mg & 1000mg Moles of CaCO3 reacting for given mass 1000 mg 100.0869 X10% mg/mol 29. 992X10-3 mol Now, 1 mole of CaCO3 requires a moles of He 9.992X10-3 mol of cacog requires ex9.ggexlo-3 mol =19.98 x103 mol of Hce Molarity of Hu z 0.15 m. 20.15 moles IL So, 0.15 mole i i present in IL HR o 19.98810-3 mol is present in 19.98X10²1 = 133x10-3 L HU 0.133 mL HU . 0.133 mL he could be neutralised by a dose of two tablets. 0-15