132 grams x (1 mol / 44 grams) = 3 moles
<span>3 moles X (22.4 L/ 1 mol) = 67.2 </span><span>L</span>
Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)

conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is 500 K.


Therefore, the value of
for the reaction is -959.1 kJ
It has to be 120g because each and every chemical equation has to satisfy the law of conservation of mass, ie sum of mass of products is always equal to the sum of masses of reactants. If reactants=120g, then products=120g
Sodium mn2 and po4 makes sodium
Answer: is letter D. Because minerals are solid.
Explanation: