Answer to Question #326532 in General Chemistry for deniz

Question

In 100 mL of 6.02 M HCl, which has a gravity of 1.10 g/ml, a strip of magnesium metal with a mass of 1.22 g dissolves. The initial temperature of the hydrochloric acid is 23.0 °C, and the end temperature of the solution is 45.5 °C. Calculate AH for the reaction under the conditions of the experiment. (The heat capacity of the calorimeter is 562 J/°C, specific heat of the finals is 4.184 /gºC.) the reaction is: Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)

Answer 1

-q=4.184(100mL)(1.10g/ml)(45.5-23.0)+562(45.5-23.0)

q=-23000J=-23.0kJ

moles Mg = 1.22 g (1 mol / 24.305) = 0.0502 moles Mg —> limiting reactant

moles HCl = 0.1 L (6.02 mol / 1 L) = 0.602 moles HCl

ΔH = -23.0kJ/0.0502 moles = -460kJ/mol