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<u>Answer:</u> The mass percent of hydrogen in methyl acetate is 8 %
<u>Explanation:</u>
The given chemical formula of methyl acetate is 
To calculate the mass percentage of hydrogen in methyl acetate, we use the equation:

Mass of hydrogen = (6 × 1) = 6 g
Mass of methyl acetate = [(3 × 12) + (6 × 1) + (2 × 16)] = 74 g
Putting values in above equation, we get:

Hence, the mass percent of hydrogen in methyl acetate is 8 %
Answer: 72.4 kJ/mol
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_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_3H_8%7D%5Ctimes%20%5CDelta%20H_%7BC_3H_8%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
![-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=-2220.1%3D%5B%283%5Ctimes%20-393.5%29%2B%284%5Ctimes%20-241.8%29%5D-%5B%285%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)

Therefore, the heat of formation of propane is 72.4 kJ/mol