given: Mg (s) + 2HCl (aq) --> MgCl2(aq) + H2 initial temp = 23.0 degrees C final temp = 32.8 degrees C weight of Mg ribbon = 0.2401 g carried out in calorimeter containing 100mL of water. Find the H in kcal/mol given: Mg (s) + 2HCl (aq) --> MgCl2(aq) + H2 initial temp = 23.0 degrees C final temp = 32.8 degrees C weight of Mg ribbon = 0.2401 g carried out in calorimeter containing 100mL of water. Find the H in kcal/mol initial temp = 23.0 degrees C final temp = 32.8 degrees C weight of Mg ribbon = 0.2401 g carried out in calorimeter containing 100mL of water. Find the H in kcal/mol
Explanation / Answer
Formula : qrxn = - s * m * T Where s is the specific heat ( for water 4.18 J / g-K) mis the mass of solution T change in temperature Data : m = 0.2401 g + 100 g ( Since D of water = 1 g /mL) = 100.2401 g T = ( 32.8 degrees C - 23 degreesC ) = 9.8 degrees C = 2.8 K qrxn = - 4.18 J / g-K * 100.2401 g *9.8 K = -4106.23 J = -4.10623 kJ No.of mols of MgO = 0.2401 g / 24.305 g/ mol = 0.0098 mols For 0.0098 mols ofMgO H is -4.10623kJ . For 1mol H is = -4.10623 kJ / 0.0098mol = -419kJ / mol = -419 / 4.184 kcal / mol = -100.14 k cal / mol