<u>Answer:</u> The enthalpy of the formation of 
is coming out to be -410.8 kJ/mol.Z
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
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%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_2H_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol](https://tex.z-dn.net/?f=81.1%3D%5B%281%5Ctimes%20%28226.7%29%7D%29%2B%284%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%2B%285%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%3D-410.8kJ%2Fmol)
Hence, the enthalpy of the formation of is coming out to be -410.8 kJ/mol.
The Enthalpy of Reaction is the adjustment in the enthalpy of a compound response that happens at a consistent weight. It is a thermodynamic unit of estimation helpful for figuring the measure of vitality per mole either discharged or created in a response.
Answer:
1.75 moles of H₂O
Solution:
The Balance Chemical Equation is as follow,
N₂H₄ + O₂ → N₂ + 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
32.04 g (1 mol) N₂H₄ reacts with = 32 g (1 mol) of O₂
So,
28 g of N₂H₄ will react with = X g of O₂
Solving for X,
X = (28 g × 32 g) ÷ 32.04 g
X = 27.96 g of O₂
It means 29 g of N₂H₄ requires 47.96 g of O₂, while we are provided with 73 g of O₂ which is in excess. Therefore, N₂H₄ is the limiting reagent and will control the yield of products.
Step 2: Calculate moles of Water produced,
According to equation,
32.04 g (1 mol) of N₂H₄ produces = 2 moles of H₂O
So,
28 g of N₂H₄ will produce = X moles of H₂O
Solving for X,
X = (28 g × 2 mol) ÷ 32.04 g
X = 1.75 moles of H₂O
Answer:
Explanation:
Hello,
In this case, given the balanced reaction:
We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:
Best regards.
