Answer: -227 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)

Therefore, the enthalpy change for
is -227 kJ.
We calculate for the amount of chromium metal in the reactant by,
= 350 x (mass of Cr2/mass of Cr2O3)
= 350 x (104/152)
= 239.47 grams
The amount of Cr metal in the product is only 213.2 grams. Thus, the percent yield.
percent yield = (213.2 grams/239.47) x 100%
= 89%
<span>Carbon Monoxide.
First, determine the relative number of moles of each element by looking up the atomic weights of carbon and oxygen
Atomic weight carbon = 12.0107
Atomic weight oxygen = 15.999
Moles of Carbon = 24.50 g / 12.0107 g/mol = 2.039847802 mol
Moles of Oxygen = 32.59 g / 15.999 g/mol = 2.037002313 mol
Given that the number of moles of both carbon and oxygen are nearly identical, it wouldn't be unreasonable to think that the empirical formula for the compound is CO which also happens to be the formula for Carbon Monoxide.</span>