How many kJ of heat are released when 15.75 g of Ba reacts completely with oxygen to form BaO?
The enthalpy for the reaction below is -1107 kJ.
2Ba(s) +O2(g) --> 2BaO(s)
Group of answer choices
20.8
63.5
114
70.3
127
Just give me the answer
Enthalpy released is 1107 for this reaction.
2Ba(s) +O2(g) --> 2BaO(s) so that is 1107/2 for this reaction:
Ba(s) + 1/2 O2(g) --> BaO(s) dH = -553.5 kJ for 137.3 kJ so for 15.75 g Ba it will be
553.5 kJ x (15.75/137.3) = ?
Check all of those numbers. Post your work if you have questions or you get stuck.
Well, it seems you're trying to heat things up with this question! Ba-ck off a bit and let me calculate it for you.
First, we need to find the moles of Ba that reacted. We can do that by dividing the mass of Ba (15.75g) by its molar mass (137.33g/mol). This gives us approximately 0.1144 moles of Ba.
Since the reaction is 2Ba(s) + O2(g) --> 2BaO(s), we can see that we need twice the amount of moles of BaO compared to Ba. Therefore, we have approximately 0.2288 moles of BaO.
Now we can use the enthalpy given (-1107 kJ) to calculate the heat released. Since the reaction produces 2 moles of BaO, we will divide the enthalpy by 2 to find the heat released per mole of BaO.
(-1107 kJ) / 2 = -553.5 kJ
So, approximately -553.5 kJ of heat are released when 15.75 g of Ba reacts completely with oxygen to form BaO.
But hey, maybe that much heat in a reaction will warm your Ba-ckyard. Just be careful not to get too close, or you might feel Ba-d!
To find out how many kJ of heat are released when 15.75 g of Ba reacts completely with oxygen to form BaO, we need to use the given enthalpy for the reaction.
1. Convert the mass of Ba to moles:
molar mass of Ba (atomic weight) = 137.33 g/mol
moles of Ba = mass of Ba / molar mass of Ba
moles of Ba = 15.75 g / 137.33 g/mol
2. Use the balanced equation to determine the stoichiometry:
From the balanced equation, we can see that 2 moles of Ba react to form 2 moles of BaO.
3. Calculate the moles of BaO formed:
moles of BaO = moles of Ba
4. Use the enthalpy change and stoichiometry to calculate the heat released:
heat released = enthalpy change × moles of BaO
heat released = -1107 kJ/mol × moles of BaO
Since 2 moles of BaO are formed for every 2 moles of Ba, the moles of BaO is equal to the moles of Ba.
Therefore, the heat released when 15.75 g of Ba reacts completely is:
heat released = -1107 kJ/mol × moles of Ba
heat released = -1107 kJ/mol × (15.75 g / 137.33 g/mol)
Now we can calculate the value:
heat released = -1107 kJ/mol × (15.75 g / 137.33 g/mol)
heat released ≈ -1107 kJ/mol × 0.1143 mol
Therefore, the approximate heat released is -1107 kJ/mol × 0.1143 mol ≈ -126.5 kJ.
The closest answer choice to this value is 127 kJ.
To calculate the amount of heat released when 15.75 g of Ba reacts completely with oxygen to form BaO, we need to use the stoichiometry of the reaction and the given enthalpy of the reaction.
First, we need to determine the number of moles of Ba (barium) involved in the reaction. This can be done by dividing the given mass of Ba (15.75 g) by its molar mass.
The molar mass of Ba is 137.33 g/mol:
15.75 g Ba * (1 mol Ba / 137.33 g Ba) = 0.11430 mol Ba
Next, using the stoichiometry of the reaction, we can determine the moles of BaO (barium oxide) produced. The stoichiometric coefficient for Ba and BaO is 2, meaning that 2 moles of Ba react to form 2 moles of BaO.
0.11430 mol Ba * (2 mol BaO / 2 mol Ba) = 0.11430 mol BaO
Since the reaction is exothermic, the enthalpy change can be calculated using the moles of BaO and the given enthalpy of the reaction (-1107 kJ).
0.11430 mol BaO * (-1107 kJ / 2 mol BaO) = -63.5 kJ
The negative sign indicates that the reaction releases heat.
Therefore, the answer is: 63.5 kJ
So, the correct answer choice is 63.5 kJ.