In: Chemistry
Consider the following reaction: 2Mg(s)+O2(g)→2MgO(s) ΔH=−1204kJ
Calculate the amount of heat transferred when 3.55 g of Mg(s) reacts at constant pressure ΔH= kJ
Given :
2 Mg (s) + O2 (g) --- > 2 MgO (s) Delta H = -1204 kJ
Mass of Mg = 3.55 g
Calculation of moles of Mg =3.55 g / molar mass of Mg
= 3.55 g / 24.305 g per mol
= 0.1523 mol
Given reaction shows that when 2 mol Mg reacts heat transferred = 1204 kJ
Lets calculate the heat for 3.55 g Mg
Heat transferred = 0.1523 mol Mg x -1204 kJ/2 mol Mg
= -91.70 kJ
Second part :
Calculation of mass of Mg
Moles of Mg = -231 kJ x 2 mol / -1204 kJ
= 0.3837 mol Mg
Mass of Mg = 0.3837 mol Mg x 24.305 g per mol
= 9.33 g
Third part :
The decomposition of MgO (s) is reverse reaction to the given.
Delta H for reverse reaction has only sign difference.
Reaction :
MgO(s) ----- >2Mg(s)+O2(g) ΔH=1204kJ
Heat absorbed = 40.6 g MgO x 1 mol MgO / molar mass of MgO ) x( 1204 kJ/2 mol)
= 40.6 g x (1 mol MgO / 40.304 g per mol) x 1204 kJ/ 2 mol MgO
= 606.4 kJ
Heat absorbed will be = 606.4 kJ