The value of ∆Ho for the reaction below is -1107 kJ:
2Ba(s) + O2(g) → 2BaO(s)
How many kJ of heat are released when 15.75 g of Ba(s) reacts completely with oxygen to form BaO(s)?
A) 35.1
B) 114
C) 70.3
D) 20.8
E) 63.5
1107 kJ x (15.75/2*atomic mass Ba) = ? kJ.
To determine the amount of heat released when 15.75 g of Ba(s) reacts completely with oxygen to form BaO(s), we need to use the equation:
q = m * ∆H
where:
q = heat released (in kJ)
m = mass of Ba(s) (in g)
∆H = change in enthalpy for the reaction (in kJ)
First, we need to calculate the number of moles of Ba(s) involved in the reaction using its molar mass. The molar mass of Ba is approximately 137.33 g/mol.
n(Ba) = mass of Ba(s) / molar mass of Ba
n(Ba) = 15.75 g / 137.33 g/mol
Next, we need to use the stoichiometry of the balanced equation to relate the number of moles of Ba(s) to the number of moles of heat released.
From the balanced equation, we can see that the stoichiometric coefficient of Ba(s) is 2 and the stoichiometric coefficient of ∆H is -1107 kJ.
Thus, we can set up the following proportion:
2 mol Ba(s) / -1107 kJ = n(Ba) mol Ba(s) / q kJ
Solving for q:
q = (n(Ba) * -1107 kJ) / 2 mol Ba(s)
Substituting the value of n(Ba):
q = ((15.75 g / 137.33 g/mol) * -1107 kJ) / 2 mol Ba(s)
Calculating this expression will give us the value of q, which represents the amount of heat released.