To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer:
5.0 moles of water per one mole of anhydrate
Explanation:
To solve this question we must find the moles of the anhydrate. The difference in mass between the dry and the anhydrate gives the mass of water. Thus, we can find the moles of water and the moles of water per mole of anhydrate:
<em>Moles Anhydrate:</em>
7.58g * (1mol / 84.32g) = 0.0899 moles XCO3
<em>Moles water:</em>
15.67g - 7.58g = 8.09g * (1mol / 18.01g) = 0.449 moles H2O
Moles of water per mole of anhydrate:
0.449 moles H2O / 0.0899 moles XCO3 =
5.0 moles of water per one mole of anhydrate
<u>Answer:</u> The amount of energy released per gram of 
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of produces -9078.57 kJ of energy.
Or,
If 
of produces -9078.57 kJ of energy
Then, 1 gram of will produce = 
of energy.
Hence, the amount of energy released per gram of is -71.92 kJ
I believe the answer is 47.25g
let me know if you need the workings i'll try and put it up
There are 7 digits from decimal to 1st digit, and it's coming from right, so exponent will be in negative 7
In short, Your Answer would be: Option C) <span>6.75 × 10-7
</span>
Hope this helps!
