given: Mg (s) + 2HCl (aq) --> MgCl 2 (aq) + H 2 initial temp = 23.0 degrees C fi

Question

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

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